There are several ways to absorb and remove unwanted surface oil. One way is to use oil-absorbing sheets or blotting paper, which can easily soak up excess oil from the skin. Another way is to apply a clay mask, which can absorb oil and impurities from the skin.
There are several ways to absorb and remove unwanted surface oil. One way is to use oil-absorbing sheets or blotting paper, which can easily soak up excess oil from the skin. Another way is to apply a clay mask, which can absorb oil and impurities from the skin. Additionally, using a toner that contains ingredients like witch hazel or salicylic acid can help to remove surface oil and keep pores clear. It's important to avoid harsh products or over-washing the skin, as this can actually stimulate the production of more oil. Instead, focus on gentle, non-drying methods to keep skin balanced and healthy.
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What product results from the reaction of CH2==CH2 with Br2?
A. CHBrCHBr
B. CH2CHBr
C. CH3CH2Br
D. CH2BrCH2Br
The correct option is:D. CH2BrCH2Br. The product that results from the reaction of CH2=CH2 (ethylene) with Br2 (bromine) is CH2BrCH2Br (1,2-dibromoethane). The reaction of CH2==CH2 with Br2 is a halogenation reaction, which involves the addition of a halogen to an unsaturated organic compound.
The product that results from this reaction is CH2BrCH2Br, which is option D. This product is formed by the addition of one Br atom to each of the carbon atoms in the double bond of ethene. The resulting molecule is a dibromoalkane, which is a type of organic compound that contains two bromine atoms attached to adjacent carbon atoms. This reaction is an example of an addition reaction, where the unsaturated organic compound undergoes a reaction with a halogen to form a saturated organic compound. In summary, the correct answer is D, CH2BrCH2Br. This reaction is an example of an addition reaction, in which the bromine atoms are added to the carbon atoms in the double bond, resulting in a single bond between the carbon atoms and a bromine atom attached to each carbon.
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what is the molar mass of a 4.10 g sample of gas exerting 1.35 atm of pressure at 325 k in a 5.00 l container? your answer should include three significant figures. provide your answer below:
The molar mass of the gas is 41.4 g/mol If a 4.10 g sample of gas exerting 1.35 atm of pressure at 325 k in a 5.00 l container.
The molar mass can be calculated using the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Rearranging the equation to solve for n, we get n = PV/RT.
First, we need to convert the pressure to units of Pascals (Pa) and the volume to units of cubic meters (m^3) to use the ideal gas law with the gas constant R = 8.31 J/(mol K):
1 atm = 101325 Pa
5.00 L = 0.00500 m^3
1.35 atm x (101325 Pa/1 atm) = 136702.5 Pa
n = (136702.5 Pa x 0.00500 m^3)/(8.31 J/(mol K) x 325 K) = 0.0991 mol
Now we can calculate the molar mass using the given mass and number of moles:
molar mass = mass/number of moles = 4.10 g/0.0991 mol = 41.4 g/mol
Therefore, the molar mass of the gas is 41.4 g/mol (to three significant figures).
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Answer:
16.2g/mol
Explanation:
[tex]Mm=\frac{mRT}{PV}[/tex]
[tex]Mm=\frac{((4.10g)(0.08206\frac{L*atm}{mol*k})(325K))}{((1.35 atm)(5.00L))}[/tex]
[tex]Mm= 16.2\frac{g}{mol}[/tex]
If the following redox reaction occured, which compound would be oxidized? Reduced?
C6H6O5 + NAD+ ---> C4H4O5 + NADH + H+
CC 9.1
In the given redox reaction: C6H6O5 + NAD+ --> C4H4O5 + NADH + H+. C6H6O5 is the compound that is being oxidized.
The process of oxidation occurs when a chemical loses electrons and becomes more positively charged, whereas reduction occurs when a compound receives electrons and becomes more negatively charged.
Since NAD+ obtains electrons and becomes more negatively charged throughout this process, it is clear that NAD+ is being reduced to NADH.
In contrast, as C6H6O5 loses electrons and becomes more positively charged, it is oxidised to C4H4O5.
Therefore, NAD+ is the chemical that is being reduced in this reaction, whilst C6H6O5 is the component that is being oxidised.
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How long will the bag last for the following order?
Potassium chloride 10 mEq
50 mL D5W
Rate: 50 mL/hr
Select one:
0.5 hour
1 hour
3/4 hour
1/4 hour
The bag containing Potassium chloride 10 mEq and 50 mL D5W given at a rate of 50 mL/hr will last for 1 hour. It is important to closely monitor patients receiving this medication and IV fluid combination to prevent adverse reactions.
Based on the order of Potassium chloride 10 mEq and 50 mL D5W to be infused at a rate of 50 mL/hr, the bag will last for 1 hour. This is because the bag contains 50 mL of solution and the rate of infusion is 50 mL/hr. This means that the entire bag will be infused over the course of 1 hour.
Potassium chloride is a medication used to treat low levels of potassium in the blood. D5W, on the other hand, is a type of intravenous fluid that contains dextrose (sugar) and water. This solution is often used to provide energy and fluids to the body.
It is important to monitor patients who receive potassium chloride and D5W to avoid complications such as hyperkalemia (high potassium levels in the blood) or hyperglycemia (high blood sugar levels). The rate of infusion should also be closely monitored to prevent fluid overload and other adverse reactions.
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1. If you place 30. 0 L of ethyl acetate (C4H8O2) in a sealed room that is 7. 25 m long, 2. 75 m wide, and 2. 75 m high, will all the ethyl acetate evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl acetate is 94. 9 torr at 25 °C, and the density of the liquid at this temperature is 0. 901 g/mL. Treat the room dimensions as exact numbers
There would be approximately 15.77 kg of ethyl acetate liquid remaining in the room. The amount of liquid remaining is significant, so it is unlikely that all the ethyl acetate would evaporate in this scenario.
When a liquid is placed in a sealed container, it will evaporate until the vapor pressure of the liquid equals the partial pressure of the vapor in the container. At this point, the liquid will be in a state of dynamic equilibrium with its vapor, and no more evaporation will occur. Therefore, the amount of ethyl acetate that evaporates in the given room depends on the vapor pressure of the liquid and the partial pressure of the vapor in the room.
First, we need to calculate the amount of ethyl acetate that would evaporate if the entire 30.0 L of the liquid were to vaporize at 25°C. To do this, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Rearranging this equation gives us [tex]\frac{n}{V} = \frac{P}{RT}[/tex], which tells us the number of moles per unit volume of a gas at a given pressure and temperature.
Using this equation, we can calculate the number of moles of ethyl acetate vapor that would be produced by 30.0 L of the liquid at 25°C and 94.9 torrs of pressure:
[tex]$n/V = \dfrac{P}{RT} = \left(\dfrac{94.9\ \text{torr}}{760\ \text{torr/atm}}\right) \left(\dfrac{1\ \text{atm}}{101.3\ \text{kPa}}\right) \left(\dfrac{30.0\ \text{L}}{0.901\ \text{g}}\right) \left(\dfrac{1\ \text{mol}}{88.11\ \text{g}}\right) \left(\dfrac{1\ \text{kPa}}{101.3\ \text{torr}}\right) = 1.08\ \text{mol/L}$[/tex]
Multiplying this by the total volume of the room gives us the total number of moles of ethyl acetate vapor that the room can hold at 25°C and 94.9 torr:
[tex]$n = \left(1.08\ \text{mol/L}\right) \left(7.25\ \text{m} \times 2.75\ \text{m} \times 2.75\ \text{m}\right) = 127.8\ \text{mol}$[/tex]
Therefore, if all 30.0 L of ethyl acetate were to evaporate, the room could hold 127.8 moles of the vapor at equilibrium.
However, we also need to consider the fact that the liquid density of ethyl acetate is 0.901 g/mL. Therefore, the mass of ethyl acetate in the room is:
m = (30.0 L) × (0.901 g/mL) = 27.03 kg
Assuming that all the liquid evaporates, the total mass of the vapor in the room at equilibrium would be:
m = n × M = (127.8 mol) × (88.11 g/mol) = 11.26 kg
Comparing this to the original mass of the liquid in the room, we can see that there is still some liquid remaining:
27.03 kg - 11.26 kg = 15.77 kg
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Determine the amount of energy absorbed by 2.00 L of gasoline as it is converted to the vapor phase at its boiling point.
The amount of energy absorbed by 2.00 L of gasoline as it is converted to the vapor phase at its boiling point is 22606 J.
What is enthalpy of vaporization?We know that;
1 mole of gasoline occupies 22.4 L
x moles of gasoline occupies 2 L
x = 0.089 moles
Then we have that the mass of the gasoline = 0.089 moles * 100 g/mol
= 8.9 g
Then;
H = mL
L = 2540 J/g for gasoline
Thus the energy that is absorbed is;
H = 8.9 g * 2540 J/g
H = 22606 J
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Approximately how many categories of chemicals are "peroxide formers?"
3
8
12
23
Approximately 3 categories of chemicals that are considered "peroxide formers."
Peroxide formers are chemicals that can form dangerous peroxides when exposed to air or light. These chemicals are typically classified into three main categories:
1. Severe peroxide hazard chemicals
2. Moderate peroxide hazard chemicals
3. Low peroxide hazard chemicals
Hence, there are about 3 categories of chemicals that are known as peroxide formers. These chemicals can form dangerous peroxides when exposed to certain conditions, and their hazard levels are classified as severe, moderate, or low.
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Which is the following statements are true with regard to displacement?
The statement that is true about displacement is D. 3 and 4 only.
What is true of displacement ?As per the halogen's reactivity series, bromine surpasses iodine in terms of its level of potency. Therefore, a reactive halogen can substitute another less reactive one from an aqueous solution of its salt.
The position of fluorine is towards the upper section of periodic table than that of chlorine, thus exhibiting more activity when compared with chlorine. A trend states that there exists an increase in oxidation ability (reactivity) of Halogens as one traverses up and across the periodic table towards right side.
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which of the following solutions is a strong electrolyte? question 27 options: a.0.10 m ammonia b.0.10 m nacl c.solid nacl d.0.10 m glucose
The solutions that is a strong electrolyte is b. 0.10 M NaCl.
When a chemical dissolves in water, it totally separates into ions, creating a high ion concentration in solution. This is referred to as a strong electrolyte. Because it totally dissociates into Na⁺ and Cl⁻ ions when it dissolves in water, NaCl (sodium chloride) is a strong electrolyte. NaCl is a good conductor of electricity in solution due to the high ion concentration.
Because it only completely dissociates into NH4₄⁺ and OH⁻ ions in solution, ammonia (NH₃) is a weak electrolyte. Because the ions are securely bound in a crystalline lattice structure and are not free to migrate in solution, solid NaCl does not conduct electricity. In solution, glucose does not separate into ions and is not an electrolyte.
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Which is the softest crystal direction in diamond?
The softest crystal direction in a diamond refers to the orientation in which the crystal structure is easiest to cleave or split. In the case of a diamond, the softest crystal direction is along its cleavage planes, specifically, the {111} planes.
Diamond has a cubic crystal structure composed of carbon atoms, with each atom bonded to four other carbon atoms in a tetrahedral arrangement. This results in a highly symmetrical and strong network of covalent bonds.
The {111} planes represent a set of crystallographic planes where the atoms are more widely spaced, and the bonding between them is weaker as compared to other directions. When force is applied along these {111} planes, the diamond is more susceptible to cleaving because the bonds between the atoms are easier to break. This softest crystal direction is essential for gem cutters, as it allows them to split and shape diamonds with greater precision, maximizing their value and aesthetic appeal.
It is important to note that while diamonds have the softest crystal direction, they are still the hardest known natural material on Earth. Their exceptional hardness is due to the strong covalent bonding between carbon atoms, which gives them a high resistance to deformation and scratching. The softest crystal direction in a diamond is only relative to the other directions within the diamond's crystal structure and does not imply that diamonds are soft materials.
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For the aqueous (Cd(CN)4] complex K, = 7. 7 x 1016 at 25 °C. Suppose equal volumes of 0. 0028 M CO(NO3), solution and 0. 16M KCN solution are mixed. Calculate the equilibrium molarity of aqueous Cd2+ ion. Round your answer to 2 significant digits. OM 1x10 Х ?
No, it is not true that the hydrogen-to-oxygen mass ratio in [tex]\mathrm{C_{24}H_{42}O_{21}}$.[/tex] is 2 to 1.
To determine the hydrogen-to-oxygen mass ratio in a compound, we need to first calculate the molar mass of the compound and then determine the ratio of the number of moles of hydrogen to the number of moles of oxygen in the compound.
The molar mass of [tex]\mathrm{C_{24}H_{42}O_{21}}$.[/tex] can be calculated by adding the molar masses of carbon, hydrogen, and oxygen atoms in the compound:
Molar mass = (24 x 12.01 g/mol) + (42 x 1.01 g/mol) + (21 x 16.00 g/mol) = 642.66 g/mol
Next, we need to determine the ratio of the number of moles of hydrogen to the number of moles of oxygen in the compound. To do this, we can divide the mass of hydrogen by its molar mass and divide the mass of oxygen by its molar mass:
Number of moles of hydrogen = (42 x 1.01 g) / (1 mol x 1.01 g/mol) = 41.58 mol
Number of moles of oxygen = (21 x 16.00 g) / (1 mol x 16.00 g/mol) = 21 mol
Therefore, the ratio of the number of moles of hydrogen to the number of moles of oxygen in [tex]\mathrm{C_{24}H_{42}O_{21}}$.[/tex] is approximately 2:1. However, the ratio of their masses is not exactly 2:1 due to the difference in their molar masses.
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Question 5
Marks: 1
Neutrons are charged, high-energy particles.
Choose one answer.
a. True
b. False
Neutrons are not charged particles, they have no electrical charge, unlike protons that are positively charged and electrons that are negatively charged. Neutrons have a neutral charge, and they do not interact with charged particles like electrons and protons, but they can interact with other particles through the strong nuclear force.
Regarding the term "high-energy," neutrons can indeed be high-energy particles in certain situations. For example, when they are emitted during a nuclear reaction, they can have a lot of kinetic energy. However, in general, neutrons have a much lower energy than other subatomic particles like protons and electrons.
In summary, neutrons are not charged particles, but they can be high-energy particles in certain contexts.
The correct answer to the question is false.
Neutrons are not charged, high-energy particles. Instead, they are neutral particles found in the nucleus of an atom, along with protons. Neutrons have no charge, meaning they are not charged particles. Protons, on the other hand, are positively charged particles found in the nucleus.
While neutrons can be involved in high-energy reactions, such as nuclear fission and fusion, they themselves are not inherently high-energy particles. High-energy particles, such as cosmic rays or particles accelerated in particle accelerators, often carry a charge and exhibit high kinetic energies.
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where do the bubbles in a glass of beer or champagne come from the bubbles are simply air bubbles resulting from the brewing process
The bubbles in a glass of beer or champagne primarily consist of carbon dioxide produced during the brewing process. The pressure change when opening a bottle or pouring the beverage allows the dissolved CO₂ to come out of the solution and form bubbles, contributing to the effervescence and mouthfeel of these popular drinks.
Bubbles in a glass of beer or champagne are an interesting phenomenon and can be explained by considering the brewing process and the physical properties of these beverages.
During the brewing process, yeast ferments sugar present in the mixture, producing alcohol and carbon dioxide (CO₂) gas as byproducts. In beer, this CO₂ is primarily responsible for the characteristic bubbles, while in champagne, secondary fermentation in the bottle generates additional CO₂. Once the beverage is bottled and sealed, CO₂ gas dissolves in the liquid under pressure.
When you pour a glass of beer or champagne, or open a bottle, the pressure decreases, allowing the dissolved CO₂ to come out of the solution and form bubbles. These bubbles are not simply air, but primarily consist of carbon dioxide produced during fermentation.
Nucleation sites, such as imperfections in the glass, dust particles, or even the tiny fibers of a cloth used to clean the glass, facilitate bubble formation by providing a surface for the CO₂ to attach and create bubbles. As these bubbles rise to the surface, they also capture other gases present in the liquid, such as nitrogen or oxygen.
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There are limits with regard to the time that hazardous waste can be stored on site for
Conditionally exempt small quantity generators
Small quantity generators
Large generators
Both (b) and (c)
Both (b) and (c) small and large quantity generators have specific time limits for storing hazardous waste on-site.
Hazardous waste refers to materials that pose significant threats to public health or the environment if improperly managed. Different types of hazardous waste generators are categorized based on the amount of waste produced. These categories are conditionally exempt small quantity generators (CESQGs), small quantity generators (SQGs), and large quantity generators (LQGs).
Conditionally exempt small quantity generators (CESQGs) produce the least amount of hazardous waste. They can store waste on-site for an indefinite period, provided that they do not exceed the accumulation limits of 1,000 kg or about 2,200 lbs of hazardous waste.
Small quantity generators (SQGs) generate more hazardous waste than CESQGs but less than LQGs. They can store waste on-site for up to 180 days or 270 days if the waste must be transported over 200 miles to a disposal facility. SQGs have an accumulation limit of 6,000 kg or about 13,200 lbs of hazardous waste on-site.
Large quantity generators (LQGs) produce the most hazardous waste. They can store waste on-site for up to 90 days and must comply with strict storage requirements. LQGs do not have a specific accumulation limit but must manage their waste properly and follow disposal regulations.
Therefore, the answer to your question is "Both (b) and (c)," referring to small quantity generators and large quantity generators. Proper hazardous waste management is crucial to minimize risks to public health and the environment.
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in the fischer esterification procedure, the ratio of alcohol to carboxylic acid was 1:2, while in the procedure using the acid chloride, the ratio of alcohol to carboxylic acid was 1:1. explain why the procedures used differing amounts of the alcohol.
In the Fischer esterification procedure, the ratio of alcohol to carboxylic acid is typically 1:1. However, when using carboxylic acids that have two or more acidic protons, the ratio of alcohol to carboxylic acid is often increased to 1:2 in order to drive the reaction towards the formation of the ester.
On the other hand, in the procedure using the acid chloride, the ratio of alcohol to carboxylic acid is typically 1:1. This is because acid chlorides are much more reactive than carboxylic acids and do not require a higher alcohol ratio to drive the reaction towards the formation of the ester. In fact, using a higher alcohol ratio may result in the formation of unwanted side products.
Furthermore, acid chlorides can react with a wider range of alcohols than carboxylic acids, including those that are less nucleophilic. This is another reason why a 1:1 alcohol to acid chloride ratio is often used in the procedure.
Overall, the differing amounts of alcohol used in the two procedures are due to the reactivity and nature of the starting materials being used.
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Using standard heats of formation, calculate the standard enthalpy change for the following reaction. 2CO(g) + 2NO(g)2CO2(g) + N2(g)
The standard enthalpy change for the given reaction is -1266.2 kJ/mol. This indicates that the reaction is exothermic, and that the formation of the products releases energy to the surroundings.
The standard enthalpy change for the given reaction can be calculated using the standard heats of formation (∆Hf) of the reactants and products. The standard heat of formation (∆Hf) is the enthalpy change that occurs when one mole of a substance is formed from its elements in their standard states at a specified temperature and pressure.The balanced chemical equation for the reaction is:2CO(g) + 2NO(g) → 2CO2(g) + N2(g)The standard heats of formation (∆Hf) for the reactants and products are:[tex]∆Hf(CO(g)) = -110.5 kJ/mol[/tex][tex]∆Hf(NO(g)) = 90.4 kJ/mol[/tex][tex]∆Hf(CO2(g)) = -393.5 kJ/mol[/tex][tex]∆Hf(N2(g)) = 0 kJ/mol[/tex]The standard enthalpy change (∆H°) for the given reaction can be calculated using the formula:[tex]∆H° = ∑n∆Hf(products) - ∑n∆Hf(reactants)[/tex]where n is the stoichiometric coefficient of each species in the balanced equation.Substituting the values of standard heats of formation into the formula, we get:[tex]∆H° = [2(-393.5 kJ/mol) + 1(0 kJ/mol)] - [2(-110.5 kJ/mol) + 2(90.4 kJ/mol)][/tex]= -1266.2 kJ/molTherefore, the standard enthalpy change for the given reaction is -1266.2 kJ/mol. This indicates that the reaction is exothermic, and that the formation of the products releases energy to the surroundings.For more such question on standard enthalpy
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1. The theory of
traits of a population change over time.
The theory of traits of a population change over time explains how people can change with respect to the strength and intensity of basic trait dimensions.
What is theory of traits?Trait theory in psychology serves as the thorry that focus on the idea that people differ whichg can be attributed to their strength as well as intensity of basic trait dimensions.
It shouuld be noted that the criteria that characterize personality traits involves the act of consistency as well as stability, along with individual differences. Natural selection give us the underswtandng of how genetic traits of a species undergo change over time.
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Conclusions
1. Compare the densities of the pre-1982 and post-1982 pennies. Using
the table to the right, state which metal is most likely used in the core
of post-1982 pennies. Explain your choice.
Metal
magnesium
aluminum
zinc
copper
silver
lead
Density
(g/cm³)
1. 74
2. 70
7. 00
8. 92
10. 50
11. 35
The pre-1982 pennies are made of an alloy of 95% copper and 5% zinc, while the post-1982 pennies have a copper-plated zinc core and are 97.5% zinc and 2.5% copper.
The densities of these metals differ, with copper being denser than zinc. The density of the pre-1982 penny is 8.94 g/cm³, while the post-1982 penny has a density of 6.87 g/cm³. This means that the metal used in the core of post-1982 pennies is most likely zinc, as its density matches that of the penny. Copper is too dense to be used in the core without significantly increasing the weight and cost of the coin. Zinc is a more cost-effective choice, and the copper plating on the outside of the penny gives it the appearance and conductivity of copper.
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What is the main hazard when working with hot plates?
Burns
Electrical shorts
Igniting flammable vapors
Vaporizing ordinarily non-volatile liquids
The main hazard when working with hot plates is the risk of a. Burns. Hot plates generate heat and can cause severe burns if users come into direct contact with the heated surface.
It is essential to handle hot plates with care, use proper safety equipment such as heat-resistant gloves, and always turn off the hot plate when not in use.
Electrical shorts are another hazard associated with hot plates. A short circuit may occur if there is a fault in the wiring or if the hot plate is used improperly. To minimize the risk of electrical shorts, ensure that the hot plate is in good working condition and follow the manufacturer's instructions for use.
Igniting flammable vapours is a potential hazard when working with hot plates, especially in laboratories or environments where flammable chemicals are present. To prevent this, always ensure that the workspace is well-ventilated, keep flammable materials away from the hot plate, and follow proper safety protocols for handling volatile substances.
Vaporizing ordinarily non-volatile liquids can also pose a hazard when working with hot plates. Heating these liquids may cause them to produce vapours that are harmful if inhaled or may even ignite. To minimize this risk, be aware of the properties of the materials being heated and follow appropriate safety measures.
In conclusion, when working with hot plates, the main hazard is a. Burns. However, other hazards such as electrical shorts, igniting flammable vapours, and vaporizing non-volatile liquids should also be considered. Always follow safety precautions and guidelines to ensure a safe working environment.
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for each atom, determine how many dots (valence electrons) should be drawn around the element symbol in the lewis structure for a lone, neutral atom.the lewis structure of an oxygen atom should have choose... dots drawn around the symbol o.the lewis structure of a calcium atom should have choose... dots drawn around the symbol ca.the lewis structure of a nitrogen atom should have choose... dots drawn around the symbol n.the lewis structure of an aluminum atom should have choose... dots drawn around the symbol al.the lewis structure of a fluorine atom should have choose... dots drawn around the symbol f.
The Lewis structure of an atom is a representation of its valence electron configuration. The number of dots drawn around the element symbol in the Lewis structure of a neutral, lone atom is equal to the number of valence electrons in that atom's outer shell.
For example, the Lewis structure of an oxygen atom should have six dots drawn around the symbol O, as oxygen has six valence electrons. Similarly, the Lewis structure of a calcium atom should have eight dots drawn around the symbol Ca, as calcium has eight valence electrons.
The Lewis structure of a nitrogen atom should have five dots drawn around the symbol N, as nitrogen has five valence electrons. The Lewis structure of an aluminum atom should have three dots drawn around the symbol Al, as aluminum has three valence electrons.
Finally, the Lewis structure of a fluorine atom should have seven dots drawn around the symbol F, as fluorine has seven valence electrons. By following the number of dots drawn around the element symbol in a Lewis structure, one can determine the number of valence electrons in the outer shell of an atom.
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What volume will 40 g of nitrogen gas (N2)
take up at room temperature and pressure?
The relative molecular mass of (N2) is 28.
Give your answer to 2 decimal places.
Hint: Remember that one mole of gas at room temperature and pressure occupies 24 dm³.
consider the pictured titration curve of an unknown acid with sodium hydroxide. titration curve of ph versus milliliters of sodium hydroxide up to 45 milliliters. with 0 milliliters of sodium hydroxide added, the ph is 1.6. with 10 milliliters of sodium hydroxide added, the curve is flat and the ph is 2.3. with 20 milliliters of sodium hydroxide added, the curve is steeply changing and the ph is 4.7. with 30 milliliters of sodium hydroxide added, the curve is flat and the ph is 7.2. with 40 milliliters of sodium hydroxide added, the curve is steeply changing and the ph is 9.7. what is the best description of the unknown acid?
Based on the information provided in the titration curve, we can make some educated guesses about the properties of the unknown acid being analyzed. First, we know that the initial pH of the solution is quite low - 1.6 - indicating that the acid is likely a strong acid. This is because strong acids are able to ionize completely in solution, leading to a high concentration of H+ ions and a low pH.
As sodium hydroxide is added to the solution, we can see that the pH gradually increases, suggesting that the acid is being neutralized by the base. At the 10 milliliter mark, the curve levels off and the pH only increases slightly. This indicates that the acid is nearly neutralized, and is likely a weak acid. Weak acids do not fully ionize in solution, and therefore require a higher volume of base to reach a neutral pH.
At the 20 milliliter mark, the curve becomes steeper and the pH increases more rapidly. This suggests that the neutralization reaction is proceeding more quickly, which could be indicative of a stronger acid. However, the pH at this point is still relatively low - 4.7 - which indicates that the acid is still quite acidic overall.
As more base is added, the pH increases more rapidly, and the curve becomes steeper again at the 40 milliliter mark. This suggests that the acid is being neutralized more quickly, indicating that it is becoming more and more basic. At the 45 milliliter mark, the pH has increased significantly, reaching a pH of 10.2.
Based on these observations, it is likely that the unknown acid is a weak acid with a relatively low initial pH. It is possible that it is a carboxylic acid or a phenol, both of which are weak acids commonly found in organic chemistry. Further analysis, such as melting point determination or infrared spectroscopy, could help to confirm the identity of the acid.
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What is the sodar of a molecule with 17 carbon atoms, 36 hydrogen atoms, and 2 oxygen atoms ?
The molecular formula of the molecule is [tex]$\mathrm{C}_{17}\mathrm{H}_{36}\mathrm{O}_{2}$[/tex].
Sodar is a term that is not commonly used in chemistry or molecular biology. However, I believe you may be referring to the term "molecular formula."
The molecular formula provides information about the types and numbers of atoms in a molecule. To determine the molecular formula of a molecule with 17 carbon atoms, 36 hydrogen atoms, and 2 oxygen atoms, we need to know the atomic masses of these elements. The atomic masses are used to calculate the total mass of the molecule, which can then be used to determine the molecular formula.
The atomic masses of carbon, hydrogen, and oxygen are approximately 12, 1, and 16, respectively. Using these values, we can calculate the total mass of the molecule:
(17 x 12) + (36 x 1) + (2 x 16) = 238
The molecular formula can be determined by dividing the total mass by the atomic mass of the empirical formula, which represents the simplest whole-number ratio of atoms in the molecule.
Thus, the molecular formula of the molecule is [tex]$\mathrm{C}_{17}\mathrm{H}_{36}\mathrm{O}_{2}$[/tex]
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Qualitative Inorganic Analysis Identification of Six Solutions Post-Lab Questions (30 Points). 1. One aspect of qualitative inorganic analysis involves the study of reactions between ions in solution. Evidence for reactions when two solutions are mixed is based on observations Name two observations one might observe to indicate that a reaction has occurred when two solutions are mixed. 2. What are spectator ions?
Formation of precipitate/colored complex, gas evolution, pH change.Ions that do not participate in a reaction and remain unchanged are called spectator ions.
At the point when two arrangements are blended during subjective inorganic investigation, there are a few perceptions that can show the event of a response. One perception is the development of an encourage or a hued complex, which can be a sign of the development of another compound because of a substance response.
One more perception is the development of gas, which can happen because of the arrival of gas from a reactant, for example, the creation of carbon dioxide gas from the response of a corrosive with a carbonate or bicarbonate.
Furthermore, an adjustment of the pH of the arrangement can likewise show the event of a response, as acidic or fundamental arrangements can influence the dissolvability and reactivity of particles.
Onlooker particles are particles that don't partake in a compound response and stay unaltered all through the response. These particles can be available in an answer when a response, however they don't influence the result of the response or respond with different particles.
At the end of the day, observer particles are not engaged with the substance condition for the response and are thusly excluded while composing a net ionic condition.
An illustration of an onlooker particle is a metal cation in an answer of an ionic compound that stays unaltered when the compound responds with a corrosive, delivering an alternate ionic compound and water. The metal cation isn't engaged with the response and is viewed as an observer particle.
Onlooker particles are essential to consider in subjective inorganic examination since they can influence the dissolvability and reactivity of particles in an answer, however they don't add to the distinguishing proof of particles present in an example.
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Suppose a 500. mL flask is filled with 0.20 mol of Br2, 1.7 mol of OCl2 and 0.60 mol of BrOCI. The following reaction becomes possible: Br₂(g) + OCl₂(g) → BrOCI(g) +BrCl(g) The equilibrium constant K for this reaction is 0.802 at the temperature of the flask. Calculate the equilibrium molarity of OCI₂. Round your answer to two decimal places.
The equilibrium molarity of OCI₂ is 2.76 M.
We can start by writing the expression for the equilibrium constant in terms of the concentrations of the reactants and products:
K = [BrOCI][BrCl] / [Br2][OCl₂]
We are given the initial moles of Br2, OCl₂, and BrOCI, but we don't know their final concentrations at equilibrium. Let's define x as the change in concentration of OCl₂ and Br₂ due to the reaction, and y as the change in concentration of BrOCI and BrCl. Then, we can write the following expressions for the equilibrium concentrations:
[Br₂] = 0.20 mol / 0.500 L = 0.40 M
[OCl₂] = (1.7 mol - x) / 0.500 L
[BrOCI] = (0.60 mol - y) / 0.500 L
[BrCl] = y / 0.500 L
We can substitute these expressions into the equilibrium constant expression and solve for x and y:
0.802 = ([0.60 - y] [y]) / ([0.40 + x] [(1.7 - x)])
Solving for x and y gives:
x = 0.32 mol
y = 0.28 mol
Now we can calculate the equilibrium concentration of OCl₂:
[OCl₂] = (1.7 mol - x) / 0.500 L = (1.7 - 0.32) / 0.500 L = 2.76 M
Rounding to two decimal places gives a final answer of 2.76 M for the equilibrium molarity of OCl₂.
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_______ WILL BOIL AT A TEMPERATURE SAME WITH 2 m sugar solution
The boiling point of the liquid depends upon the pressure of the surrounding. A liquid at high pressure has a higher boiling point than the boiling point at normal atmospheric pressure. Here any solution with 2m concentration will boil at a same temperature.
The temperature at which the vapor pressure of the liquid becomes equal to the atmospheric pressure of the liquids environment is the boiling point. It is at this temperature, the liquid is converted into a vapour.
Since on increasing the number of moles the molality also increases and is directly proportional to an elevation in boiling point. So any solution with 2m concentration boil at same temperature.
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Which property describes observable characteristics of matter like color?
Chemical Property
Physical Property
Reactivity
Sublimation
Pls help will give brainliest and 100 points.
Define electronegativity in Your own words.
Answer: Electronegativity is the tendency of an atom to attract electrons in a molecule
Explanation:
What is Delta S greater than 0?
Delta S greater than 0 refers to a positive change in entropy, where the entropy of a system increases.
This means that the system becomes more disordered or random, and there is a greater number of possible arrangements or configurations of its particles. This can occur due to various factors, such as an increase in temperature, a phase transition, mixing of different substances, or chemical reactions that produce more products than reactants. A positive Delta S value is important in thermodynamics, as it indicates the direction of spontaneous processes that are favorable in terms of energy and entropy.
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for a given fluorophore, select the choice that correctly lists the processes of fluorescence, absorption, and phosphorescence in order from shortest to longest wavelength. fluorescence < phosphorescence < absorption absorption < phosphorescence < fluorescence phosphorescence < fluorescence < absorption absorption < fluorescence < phosphorescence absorption < fluorescence
For a given fluorophore, absorption < fluorescence < phosphorescence is correct.
Fluorescence, absorption, and phosphorescence are all related to the way light interacts with matter.
Absorption occurs when a molecule absorbs a photon of light, which causes an electron to jump to a higher energy level. This process occurs at a specific wavelength, which is unique to each molecule. The absorbed energy is usually converted into heat or used to drive a chemical reaction.
Fluorescence occurs when a molecule that has been excited by absorbing light emits a photon of light as it returns to its ground state. This process occurs at a longer wavelength than the absorbed light, and the emitted light is usually of a different color than the absorbed light. Fluorescence occurs quickly, typically within nanoseconds of the molecule being excited.
Phosphorescence is a type of delayed fluorescence that occurs when a molecule remains in an excited state for a longer period of time, typically microseconds to seconds. During this time, the molecule can emit light as it returns to its ground state. Phosphorescence occurs at an even longer wavelength than fluorescence.
The order of these processes in terms of wavelength is absorption < fluorescence < phosphorescence. When a molecule absorbs light, it does so at a specific wavelength. When it fluoresces, it emits light at a longer wavelength. When it phosphoresces, it emits light at an even longer wavelength. Therefore, the wavelength increases as we move from absorption to fluorescence to phosphorescence.
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