Everyone is aware that the first billion years of Earth's existence were marked by intense volcanic activity. Because of the circumstances, there were so many gases in our atmosphere that life could not exist there (Very similar like the atmosphere of Mars & Venus of today).
Carbon dioxide was most prevalent, whereas oxygen was scarce. Next, the modifications that affected the ratio of gases were:
1) To increase oxygen levels. "Photosynthesis by Plants" was the sole and most significant source of increasing oxygen.
2) In order to reduce carbon dioxide. - I Absorbable by plants for the photosynthesis process.
(ii) The majority of the CO2 is dissolved in the oceans, and a large portion of it is trapped in sedimentary rocks like limestone and fossil fuels.
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Think of a materials that can produce sound and it can be changed into another form of energy that you would like to create. Draw and illustrate the design of your inverntion. Describe your invention and how it can help people
One material that can produce sound and be converted into another form of energy is piezoelectric material. Piezoelectric materials have an electrical charge when subjected to mechanical stress.
What is Piezoelectric material?Piezoelectric material produces an electrical charge when subjected to mechanical stress, such as vibrations or pressure. This means that they can convert sound waves into electrical energy, which can then be used to power other devices.
What is the use of piezoelectric material?One potential application of this technology is in developing piezoelectric generators that can be installed in public spaces, such as parks or city streets, to capture the energy from ambient sound waves and convert it into electricity. The electricity generated by these devices could be used to power streetlights, charging stations for mobile devices, or other public amenities.
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6) What is the volume occupied by 40 g of oxygen gas at STP?
Answer:
28 liters of O2
Explanation:
A marvelous byproduct of establishing a STP (Standard Temperature and Pressure) is a conversion factor that 1) seems impossible, and 2) saves a lot of time. That factor is 22.4 Liters/mole gas. At STP ALL gases occupy 22.4 liters per mole of that gas. Hydrogen, argon, methane, etc. all will occupy 22.4 liters at 1 atm pressure and 0°C.
That made no sense to me, but after realizing the amount of time it can save answering question as this, I'm a firm believer.
Without this conversion factor, the way to answer the question is to use the ideal gas law: PV=nRT, where n is the number of moles and R is the universal gas constant. Not a super difficult equation, but looking up the gas constant and making sure all the units match and cancel can take time and requires focus.
But the conversion factor is straightforward. We are given 40 g of O2. Let's coivert the 40g into moles O2. Divide the mass by the molar mass of O2 (32 g/mole):
(40g O2)/(32 g/mole O2) = 1.25 moles of O2
Now is when the ideal gas law might be employed. We have P, T, n, and R, so plug them is and calculate volume. This is done below, just as an illustration.
But since we are at STP, we can use the handy coversion factor to determine the volume of O2:
(1.25 moles O2)*(22.4L/mole O2) = 28.0 liters of O2
In contrast, the ideal gas law calculation would look like this:
Rearrange to isolate V, the unknown:
V = (nRT)/P
Enter the data, ensuring the correct gas constant is used for the units provided:
V = [(1.25 moles)*(0.0820575L⋅atm⋅K^1⋅mol^1)(273.15K)]/(1atm)
Ouch
V = 28.0 liters
Same answer, more effort.
Reasons why recycling scrap copper is more sustainable than extracting copper from copper ores
Answer:
Recycling scrap copper is more sustainable than extracting copper from copper ores due to the following reasons:
Reduces energy consumption: Recycling scrap copper requires less energy compared to mining, extracting, and processing copper ores. This results in a reduction in greenhouse gas emissions and a decrease in energy consumption.
Conservation of natural resources: Recycling scrap copper conserves natural resources because it reduces the need for mining copper ores. Mining copper ores often leads to the destruction of ecosystems, soil erosion, and the emission of harmful pollutants into the environment.
Reduction in landfill waste: Recycling scrap copper helps to reduce the amount of waste in landfills. Copper is a valuable resource, and recycling it ensures that it is reused instead of ending up in landfills.
Cost-effective: Recycling scrap copper is often cheaper than extracting copper from copper ores. This is because recycling eliminates the cost of mining, extracting, and processing copper ores.
Preservation of water resources: Extracting copper from copper ores requires large amounts of water, which can lead to water shortages in some regions. Recycling scrap copper requires less water, thus helping to preserve water resources.
Overall, recycling scrap copper is more sustainable than extracting copper from copper ores because it reduces energy consumption, conserves natural resources, reduces landfill waste, is cost-effective, and helps to preserve water resources.
Explanation:
You are provided with three unlabelled tubes containing colourless liquids. One contains an alkene, one an alkane and one water. Describe test-tube reactions that you could carry out to identify each substance. (6 marks)
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Explanation:
To identify the contents of the three unlabelled test tubes, you could carry out the following test-tube reactions:
To identify the alkene:
Add bromine water (dilute aqueous solution of bromine) dropwise to each test tube. The alkene will react with the bromine water and decolourize it, giving a clear solution. The alkane and water test tubes will not show any reaction.
To identify the alkane:
Add a few drops of a concentrated sulfuric acid to each test tube. The alkane will not react with the sulfuric acid, and there will be no observable reaction. The alkene and water test tubes will show no reaction or some reaction, but not as significant as the alkane test tube.
To identify water:
Add a few drops of anhydrous copper (II) sulfate to each test tube. The anhydrous copper (II) sulfate will react with any water present in the test tube and turn blue. The alkane and alkene test tubes will not show any reaction with copper (II) sulfate.
By performing these test-tube reactions, you should be able to identify which test tube contains an alkene, an alkane, and water.
at 800 mm Hg, a gas has a volume of 760 L, what is its volume at standard pressure?
At 800 mm Hg, a gas has a volume of 760 L, its volume at standard pressure is 800 L. This is according to Boyle's Law.
What is Boyle's Law?A gas law known as Boyle's law asserts that a gas's pressure is inversely proportional to its volume when it is held at a fixed temperature and of a given mass. To put it another way, as long as the temperature and volume of the gas remain constant, the pressure and volume of the gas are inversely proportional to one another. The Anglo-Irish chemist Robert Boyle proposed Boyle's law in the year 1662.
According to Boyle's law, a gas's pressure will change if its volume changes while remaining the same quantity and temperature. To put it another way, the ratio of a gas's beginning pressure to its initial volume is equal to the ratio of the gas' final pressure to its final volume (at constant temperature and number of moles).
Using the formula:
P₁V₁ = P₂V₂
Substituting the values and solving for V₂:
V₂= 800L
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note how the air pressure changes as you move station b towards the center of the low pressure system
High pressure centers will typically bring fair weather with little to no precipitation to cities. In contrast, a low pressure center is typically accompanied by clouds, precipitation, and anticlockwise winds.
What transpires to air heading in the direction of a low pressure centre?A low pressure system has lower pressure at its center than the surroundings around it. When winds blow in the direction of a low pressure area, the air where they meet rises in the atmosphere. The water vapor in the air condenses as it rises, creating clouds and frequently precipitation.
What way does air flow in a system with low pressure?Winds at the surface move clockwise (anticyclonically) around high pressure and anticlockwise (cyclonically) around low pressure. These systems' actual pressure can be calculated in millibars or inches of mercury, for example, 30.10. (e.g., 1004 mb).
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7. Which of the following is an element? a) Sugar b) Copper c) Salt d) Water
Answer: B copper
Explanation:
all the rest are compounds
Why is salt mixed in with ice in an ice cream maker? Select all that apply
A) the salt lowers the freezing point of the ice-water mixture.
B) the salt raises the freezing point of the ice water mixture.
C) The ice cream freezes faster.
D) The salt prevents the ice cream from freezing ununiformly.
Answer:
A) The salt lowers the freezing point of the ice-water mixture.
Explanation:
Ice cream freezes at a lower temperature than water because the presence of sugar and fat prevents the formation of ice crystals, necessitating a lower temperature to achieve freezing. Ice by itself cannot adequately chill the ice cream base because it will melt before the base has cooled enough. Water's freezing/melting point can be lowered with the use of salt. Just consider the effects salt has on the roadways when it melts ice. Similar to the last example, adding salt to the ice that surrounds the ice cream base creates a temperature that is chilly enough to prevent the ice cream within from melting completely before it has a chance to thicken and freeze.
Thanks,
Eddie.
What is the formula for lead (IV) oxide?
Answer:
Explanation:
The formula for lead (IV) oxide is PbO2
Students set up a controlled experiment. They put five
crickets in each of three identical containers. They set one
container to 15°C, one to 20°C, and one to 25°C. Then they
count the number of cricket chirps in each container during
1 hour.
What are two constants in this experiment?
A. The temperature of each container
B. The number of crickets in each container
C. The number of cricket chirps
D. The kind of containers used
Explanation:
Constants do not vary ..... they stay 'constant'
so
A .....no ...different temps
B.....constant
C ......no the # chirps vary
D.....constant
ASAP PLEASE! Mostly just need the data and conclusion answers please!!!
Electromagnetic Spectrum Lab Report
Instructions: In this virtual lab, you will use a virtual spectrometer to analyze astronomical bodies in space. Record your hypothesis and spectrometric results in the lab report below. You will submit your completed report to your instructor.
Name and Title:
Include your name, instructor's name, date, and name of lab.
Objectives(s):
In your own words, what is the purpose of this lab?
Hypothesis:
In this section, please include the predictions you developed during your lab activity. These statements reflect your predicted outcomes for the experiment.
Procedure:
The materials and procedures are listed in your virtual lab. You do not need to repeat them here. However, you should note if you experienced any errors or other factors that might affect your outcome. Using your summary questions at the end of your virtual lab activity, please clearly define the dependent and independent variables of the experiment.
Data:
Record the elements present in each unknown astronomical object. Be sure to indicate “yes” or “no” for each element.
Hydrogen Helium Lithium Sodium Carbon Nitrogen
Moon One
Moon Two
Planet One
Planet Two
Conclusion:
Your conclusion will include a summary of the lab results and an interpretation of the results. Please answer all questions in complete sentences using your own words.
Using two to three sentences, summarize what you investigated and observed in this lab.
Astronomers use a wide variety of technology to explore space and the electromagnetic spectrum; why do you believe it is essential to use many types of equipment when studying space?
If carbon was the most common element found in the moons and planets, what element is missing that would make them similar to Earth? Explain why. (Hint: Think about the carbon cycle.)
We know that the electromagnetic spectrum uses wavelengths and frequencies to determine a lot about outer space. How does it help us find out the make-up of stars?
Why might it be useful to determine the elements that a planet or moon is made up of?
Answer: This lab's goal is to investigate the absorbance patterns created by recently discovered moons and planets.
The first moon consists of Lithium and carbon, the second moon consists of sodium and nitrogen. Moving onto the planets, the first planet consists of hydrogen and carbon, and lastly, the second planet is consistent with helium and carbon.
How to explain the lab report
The theory was right; there have been no flaws in the outcome. The astronomical item observed by the spectrometer is the independent variable. The spectrum of any astronomical object is the dependent variable.
Space consists of bodies with different types of the electromagnetic spectrum. This includes high-energy bodies emitting radiation in short wavelengths and extremely short wavelengths such as in UV spectrum, X rays, and gamma rays. Conversely, other bodies might be emitting radiations in the longer wavelengths such as Microwaves and Radio waves.
The element missing from the moons and the planets would be Oxygen. It is to be remembered that Oxygen forms the base of the sustenance of life forms on Earth and forms an indispensable part of the carbon cycle. In the absence of oxygen, these planets and moons remain lifeless.
Stars emit heat and light. Along with the heat and light, radiations are emitted by the star. These radiations travel outward from stars and work as the signature of the stars. By analyzing the radiations from the stars, scientists back on Earth could deduce the physical conditions in the heart of a star including its constitution, temperature, and surface conditions.
The knowledge of the constitution of the elements making up the moon or planet is necessary to ascertain the life-sustaining capability of the same.
Answer: This lab's goal is to investigate the absorbance patterns created by recently discovered moons and planets.
The first moon consists of Lithium and carbon, the second moon consists of sodium and nitrogen. Moving onto the planets, the first planet consists of hydrogen and carbon, and lastly, the second planet is consistent with helium and carbon.
How to explain the lab report
The theory was right; there have been no flaws in the outcome. The astronomical item observed by the spectrometer is the independent variable. The spectrum of any astronomical object is the dependent variable.
Space consists of bodies with different types of the electromagnetic spectrum. This includes high-energy bodies emitting radiation in short wavelengths and extremely short wavelengths such as in UV spectrum, X rays, and gamma rays. Conversely, other bodies might be emitting radiations in the longer wavelengths such as Microwaves and Radio waves.
The element missing from the moons and the planets would be Oxygen. It is to be remembered that Oxygen forms the base of the sustenance of life forms on Earth and forms an indispensable part of the carbon cycle. In the absence of oxygen, these planets and moons remain lifeless.
Stars emit heat and light. Along with the heat and light, radiations are emitted by the star. These radiations travel outward from stars and work as the signature of the stars. By analyzing the radiations from the stars, scientists back on Earth could deduce the physical conditions in the heart of a star including its constitution, temperature, and surface conditions.
The knowledge of the constitution of the elements making up the moon or planet is necessary to ascertain the life-sustaining capability of the same.
Which set of products is correct for
this double replacement reaction?
AgNO3(aq) + Na₂CO3(aq) -
Remember that each formula must
have a balanced charge.
AgNO3(aq) + Na2CO3(aq) → Ag2CO3(s) + 2NaNO3(aq)
AgNO3 and Na2CO3 do they produce a precipitate?Response and justification The chemical equation below is a representation of the reaction between Na2CO3 N a 2 C O3 and AgNO3 A g N O3. When the aqueous solutions of Na2CO3, Na2CO3, and AgNO3, AgNO3, are combined, precipitation will occur.
What occurs when Na2CO3 and AgNO3 are combined?Both Na2CO3 and AgNO3 are strong electrolytes and will totally dissociate. The net ionic equation will divide them up as a result. The twofold displacement reaction is Na2CO3 + AgNO3. In the reaction, the positions of the Na in Na2CO3 and the Ag in AgNO3 change.
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How pure do you think the water that you use every day is? Explain.
Both direct and indirect uses of water are possible.While bathing, drinking, & cooking are examples of direct uses of water, indirect uses include the production of steel for automobiles and the processing of wood to manufacture paper.
What role does water play in our daily lives, in your opinion?It's critical to drink enough water each day for your health.By staying hydrated, you can avoid being dehydrated, a condition that really can impair your memory, alter your mood, cause your system to overheat, & result in constipation & kidney stones.
What function does water serve in your day-to-day life?Water is a crucial element of the majority all biological processes because it acts as a solvent.Using water as a solvent, food is absorbed by our bodies as solutions.Furthermore, perspiration and urine are used to eliminate several wastes as solutions.
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how many grams of solid ammonium chloride must be dissolved in 1.00 l of 0.200 m ammonia to make a buffer of ph 8.50? (kb of ammonia is 1.8x10-5).
To make a buffer of pH 8.50 with 1.00 L of 0.200 M ammonia, you need to dissolve 5.6 g of solid ammonium chloride.
This can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log ( [salt]/[acid] )
8.50 = 9.25 + log ( [NH4Cl]/[NH3] )
[NH4Cl]/[NH3] = 10^(pH-pKa)
[NH4Cl]/[NH3] = 10^(8.50-9.25)
[NH4Cl]/[NH3] = 10^(-0.75)
[NH4Cl] = [NH3] * 10^(-0.75)
[NH4Cl] = 0.200 M * 1.00 L * 10^(-0.75)
[NH4Cl] = 0.158 M
Mass of NH4Cl = 0.158 M * (5.6 g/mol) * 1.00 L
Mass of NH4Cl = 5.6 g
Hence, To make a buffer of pH 8.50 with 1.00 L of 0.200 M ammonia, you need to dissolve 5.6 g of solid ammonium chloride.
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A 550. 0 mL sample of gas at 40. 0 °C and 895 torr is transferred to a second vessel where the temperature is 0. 0 °C and the pressure is 745 torr. What is the volume of the second vessel?
The volume of the second vessel is approximately 322.7 mL.
To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:
(P₁ × V₁)/T₁ = (P₂ × V₂)/T₂
where P₁, V₁, and T₁ are the initial pressure, volume, and temperature, respectively, and P₂ and V₂ are the final pressure and volume, respectively.
First, we need to convert the temperatures to Kelvin, which is done by adding 273.15 to the Celsius temperature:
T₁ = 40.0 + 273.15 = 313.15 K
T₂ = 0.0 + 273.15 = 273.15 K
Next, we can plug in the given values and solve for V₂:
(P₁ × V₁)/T₁ = (P₂ × V₂)/T₂
(895 torr × 550.0 mL)/313.15 K = (745 torr × V₂)/273.15 K
Simplifying and solving for V₂, we get:
V₂ = (745 torr × 550.0 mL × 313.15 K) / (895 torr × 273.15 K)
≈ 322.7 mL
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The compound iron oxide can exist with either iron(II) ions or iron(III) ions. Conduct Internet research to learn about the differences between iron(II) oxide and iron(III) oxide. Give the chemical formula for each compound. Describe their appearance and uses. Based on your findings, are these two forms of the same compound, or are they two completely different compounds?
Chemical fοrmula οf Irοn(II) οxide is FeO and Irοn(III) οxide is Fe2O3.(FeO) is a black οr brοwn pοwder that is used as a catalyst, reducing agent, and pigment.
What are cοmpοund?A cοmpοund is a substance made οf twο οr mοre elements that are chemically bοnded tοgether. Cοmpοunds are fοrmed when atοms οf different elements share electrοns, fοrming chemical bοnds.
It is alsο used in the manufacturing οf ferrοus allοys, steel, and glass. Irοn(III) οxide (Fe₂O₃) is alsο a black οr red pοwder, but with a different chemical cοmpοsitiοn. It is used as a pigment and abrasive in paints and cοatings, a catalyst in the synthesis οf ammοnia, and a cοlοring agent in ceramics and plastics.
These twο fοrms οf irοn οxide are nοt the same cοmpοund, as they have different chemical fοrmulas and uses.
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35. 2 mL of gas starts out at STP; but, conditions change to 460 mm Hg atm and 193 K. Calculate volume?
The combined gas law, which relates the initial and final conditions of pressure, volume, and temperature of a gas, can be used to solve this issue.
Calculate volume?(P1V1) / (T1) equals (P2V2) / (T2) where: The initial pressure and temperature at STP are, respectively, P1 and T1 (1 atm and 273 K) The initial volume, or V1, is specified as 2 mL. P2 and T2 represent the final pressure and temperature in the new conditions, respectively, and V2 represents the volume we need to calculate. The given pressure can first be converted to atm: P2 = 460 mm Hg is equal to 460/760 atm, or 0.6053 atm. Additionally, we can change the stated temperature to Kelvin: T₂ = 193 K. Now that the values have been entered, we can solve for V2 using the combined gas law: (0.6053 atm)(V2) / (1 atm)(2 mL)/273 K (193 K), ((1 atm)(2 mL)(193 K)] = V2. / [(273 K)(0.6053 atm)] V₂ ≈ 1.14 mL. As a result, the gas's final volume under the new circumstances is roughly 1.14 mL.
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Calculate the mass of sodium tetraoxosulphate VI formed when 0. 5 mol of sodium hydroxide with tetraoxosulphate VI acid
The mass of sodium Tetraoxosulphate formed when 0.5 mole of sodium hydroxide react with Tetraoxosulphate acid is 35.5 grams.
0.5 m of NaOH denotes the amount of the solvent that is dissolved in 0.5 mole (20.0 g) of NaOH. 0.5 moles of NaOH (20.0 g) are dissolved in 1000 millilitres of the solution, or 0.5 M of NaOH.
Tetraoxosulphate VI acid is produced commercially using the Contact process. The Contact procedure entails the following steps. In order to purge the sulphur (IV) oxide generated of contaminants and dust that can poison the catalyst, it is combined with extra air and sent through an electric chamber.
Strong acid known as tetraoxosulphate may be used to produce synthetic colours, extract metals, serve as the electrolyte in lead-acid storage batteries, and produce synthetic and natural fibres.
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Complete question:
Calculate the mass of sodium tetraoxosulphate (vi) formed when 0.5 mole of sodium hydroxide react with tetraoxosulphate (vi) acid. (Na=23, O=16, S=32, H=1
Calculate the reacting mass of Propone. (C₃H₅) when it is completely burnt in C₃H₈ + O₂ → CO₂ + H₂O
Answer: First we need to balance the elements other than oxygen and hydrogen on both sides. C3H8+O2→3CO2+H2O Now, balance the number of hydrogens on both sides:.
Explanation:
Answer:To calculate the reacting mass of propane (C₃H₈) when it is completely burnt in the given reaction, we need to use the balanced chemical equation and the molar masses of the reactants and products.
The balanced chemical equation for the combustion of propane is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
From the equation, we can see that one mole of propane (C₃H₈) reacts with five moles of oxygen (O₂) to produce three moles of carbon dioxide (CO₂) and four moles of water (H₂O). The molar masses of the compounds involved in the reaction are:
Propane (C₃H₈): 44.1 g/mol
Oxygen (O₂): 32.0 g/mol
Carbon dioxide (CO₂): 44.0 g/mol
Water (H₂O): 18.0 g/mol
To calculate the reacting mass of propane, we need to know the mass of oxygen required to react completely with one mole of propane. From the balanced chemical equation, we can see that 5 moles of O₂ are required to react with 1 mole of C₃H₈. Therefore, the mass of oxygen required is:
mass of O₂ = 5 × molar mass of O₂ = 5 × 32.0 g/mol = 160.0 g/mol
Now, we can use the mass of oxygen required to calculate the reacting mass of propane. If 160.0 g of oxygen reacts with one mole of propane, then the mass of propane that would react completely with 160.0 g of oxygen is:
reacting mass of C₃H₈ = (1 mole of C₃H₈ × 44.1 g/mol) / (5 moles of O₂ × 32.0 g/mol) = 0.275 g
Therefore, the reacting mass of propane (C₃H₈) when it is completely burnt in the given reaction is 0.275 g.
Explanation: May i get brainliest :D
To earn full credit for your answers, you must show the appropriate formula, the correct substitutions , and your answer including the correct units
Golden City has 4.2 x106 people and has 5.7 x 103 births. What is the birth rate?
The birth rate is obtained to be 1.36 births per year
How do you find the birth rate?The birth rate is typically calculated as the number of births per 1,000 people in a given population over a specific period of time.
The formula that we can use so as to obtain the birth rate can be given as;
Birth rate = (Number of births / Population) x 1,000
We can find the birth rate from;
5.7 x 10^3 / 4.2 x10^6 x 1,000
1.36 births per year
This calculation provides an estimate of the number of births per 1,000 people in the population over a specific time period.
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MATH PROBLEM! A gas at STP with a volume of 13.5 liters is heated to 300 K. What is the new volume of the gas, assuming the pressure remains constant? (5 points)
The new volume of the gas at 300 K is approximately 14.8 liters.
What is the new volume of the gas?To solve this problem, we can use the following formula:
V1/T1 = V2/T2
Where;
V1 is the initial volume, T1 is the initial temperature in Kelvin, V2 is the final volume, and T2 is the final temperature in Kelvin. Since the pressure remains constant, we can use this formula to find the new volume of the gas.At STP, the temperature is 273 K, so we have:
V1/T1 = V2/T2
13.5/273 = V2/300
Solving for the new volume of the gas, V2, we get:
V2 = (13.5/273) * 300
V2 = 14.8 liters (rounded to one decimal place)
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indicate whether each statement is true or false. (a) if you compare two reactions with similar collision factors, the one with the larger activation energy will be faster. (b) a reaction that has a small rate constant must have a small frequency factor. (c) increasing the reaction temperature increases the fraction of successful collisions between reactants.
First two statements are false while the third one is true. If two reactions have similar collision factors, the one with a smaller activation energy will be faster. The rate constant depends on both the frequency factor and the activation energy.
(a) False. If two reactions have similar collision factors, the one with a smaller activation energy will be faster. Reactions with lower activation energies require less energy for the reactants to reach the transition state and become activated.
(b) False. The rate constant depends on both the frequency factor and the activation energy. Therefore, a reaction with a small rate constant could have a large frequency factor and a large activation energy.
(c) True. Increasing the temperature of a reaction increases the kinetic energy of the reactant molecules, causing them to move faster and collide more frequently. This leads to an increase in the fraction of successful collisions between reactants.
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Is my answer for 73 right?
Answer:
yes it right. well done try more questions.
igcse 2023 past papers edexcel, does anyone have the papers..? pls upload if do, i want chem, human bio, maths and physics,pls if u hv a big help
Unfortunately, the Edexcel IGCSE 2023 past papers are not yet available. We suggest that you check the Edexcel website periodically for updates.
What is available ?
There are many types of items available for purchase. Consumers can find a wide variety of goods, from clothing and electronics to food and furniture. Online retailers offer a convenient way to shop, with many items available for delivery or pick up. Consumers may also find items from local stores or through online auctions. Additionally, products can be purchased through subscription services, such as meal kits or beauty boxes. Finally, services such as home cleaning, pet sitting, and tutoring can also be purchased.
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What is the name of the formula NI3 PO4?
Nickel (I) phosphate
Nickel (III) Phosphate
Answer:
Nickel (I) Phosphate
Explanation:
PO4 is a -3 ion it will need 3 Ni(I) ion, which is -1 ion. Therefore it is Ni3PO4.
draw and label the diagram of parts of the battery and illustrate the flow of electric current.
Conventional direction is from positive terminal to negative terminal and this is the direction of the electric field within the wire.
What is the flow of current in a battery?Single cell or other power source is represented by long and short parallel line. Collection of cells or battery is represented by collection of long and short parallel lines.
In electrochemical cell, higher positive potential is cathode, therefore conventional current direction is from cathode to anode through conductor (metallic path) and from anode to the cathode in electrolyte
In an electric circuit, battery is charge pump which pumps charge through battery from location of low electric potential energy (the - terminal) to location of high electric potential energy (the + terminal). The battery does not supply electric charge as the charge is already in the wire.
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The Lewis dot diagram for an unidentified element is shown below.
Which element could X represent?
A. iron
B. chlorine
C. silicon
D. oxygen
Answer:
D. Oxygen.Explanation:
What is Lewis dot diagram?
Lewis dot diagram also known as electron -dot diagram represents the number of valence electron of atoms within a molecule.
Gilbert Lewis is the creator of the Lewis dot structure.
Lewis proposed a model for the role of electrons, particularly valence electrons, in 1916. This included the creation of electron dot diagrams or Lewis structures.
Electronic configuration of oxygen is [tex]\rm \underline{ 1s^2 \: 2s^2 \: 2p^4}[/tex]
Since oxygen has six valence electrons so, six electrons are drawn around oxygen atom.
In the given question there are also 6 electrons around the element X.
Therefore, The required answer will be Oxygen.
Which of the following would you expect to see written as separated ions in an ionic equation?
H2(g)
H2O(l)
H2S(g)
HCl(aq)
HCl(aq) would be expected to see written as separated ions in an ionic equation.
When HCl is dissolved in water, it ionizes completely to form H+ and Cl- ions, which are free to move around in solution. Therefore, in an ionic equation, HCl(aq) would be written as H+(aq) + Cl-(aq).
In contrast, H2(g), H2O(l), and H2S(g) are covalent compounds that do not ionize in water, so they would not be written as separated ions in an ionic equation.
How will the rate of reaction -A+B→AB change if A and gases increase 3-3 times? V=
The rate of the reaction will increase if the concentration of A and gases increase.
Rate of reactionsIf the concentration of reactant A and gas increases by 3 times, the rate of the reaction -A+B→AB will also increase.
This is because an increase in the concentration of reactant A means that there will be more A molecules colliding with B molecules, leading to an increase in the number of successful collisions and the formation of more AB product molecules.
Similarly, an increase in gas concentration will increase the number of gas molecules colliding with reactant molecules, which can also increase the rate of the reaction.
Therefore, the rate constant, k, of the reaction will also increase, resulting in an overall increase in the rate of the reaction.
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The reaction of 7.1 grams of chlorine with
excess fluorine produced 7.4 grams of ClF3.
What percent yield of ClF3 was obtained?
Answer in units of %
The percent yield of [tex]ClF_3[/tex] was obtained is 39.97% when the reaction of 7.1 grams of chlorine with excess fluorine produced 7.4 grams of [tex]ClF_3[/tex].
Given the mass of chlorine = 7.1g
The mass of chlorine trifluoride produced = 7.4g
Molar mass of chlorine used = 70.906g/mole
Molar mass of [tex]ClF_3[/tex] used = 92.448g/mole
The reaction of chlorine with fluorine to produce chlorine tri fluoride is as:
[tex]Cl_2 + 3F_2 -- > 2ClF_3[/tex]
1 mole of [tex]Cl_2[/tex] is used to produce 2 moles of [tex]ClF_3[/tex]
mass of [tex]Cl_2[/tex] used = number of moles x molar mass = 1 * 70.906= 70.906g
Mass of [tex]ClF_3[/tex] used = 2 * 92.448 = 184.896g
Here for 70.906g of [tex]Cl_2[/tex], 184.896g of [tex]ClF_3[/tex] is produced.
Then for 7.1g of [tex]Cl_2[/tex] = 184.896 * 7.1/70.906 = 18.51g of is produced.
Theoretical Yield of [tex]ClF_3[/tex] = 18.51g
Actual Yield = 7.4g of [tex]ClF_3[/tex]
Percent Yield =[tex](7.4/18.51 )*100 = 39.97[/tex]% yield of [tex]ClF_3[/tex]
Therefore, 39.97% yield of [tex]ClF_3[/tex] was obtained.
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