Answer
0.0779 g SO₃
Procedure
To solve this question, you will need to use the Avogadro's number equivalent to 6.02214 x 10^23 molecules/ mol, and then the molecular weight in grams per mole for the SO₃.
[tex]5.86\times10^{20}\text{ molecules SO}_3\frac{1\text{ mol SO}_3}{6.02214\text{ x 10}^{23}\text{molecules SO}_3}\frac{80.06\text{ g SO}_3}{1\text{ mol SO}_3}=0.0779\text{ g SO}_3[/tex]If an acid dissosciates, what is happening?A. It is combining with a base in a neutralization reactionB. it is involved in a chemical reactionC. the concentration is decreasingD. it is being pulled apart by solvent molecules
In a solution, when we have an acid being dissociated, what is actually happening is that the acid is being pulled apart by solvent molecules, becoming two ions, the proton H+ and the conjugated base, this is what causes the solution to be acidic, and this is also how the strength of the acid is measured. Therefore the best answer is letter D
4. An ice cube (25 g) is at -8.0°C. How much energy is required to take it to the
melting point, 0 °C? Heat capacity (c) for solid water is 2.10 J/g C
the heat energy required to take it to the melting point is 420J.
What is specific heat capacity?The heat capacity, abbreviated Cp, is the amount of heat needed to elevate a mole of a substance's heat content by precisely one degree Celsius.
A material has more thermal energy the hotter it is, says basic thermodynamics. Additionally, when a chemical is present in greater concentrations at a particular temperature, it will have a higher total thermal energy.
Mathematically,
Q = mc∆T
Where,
Q = quantity of heat absorbed by a body
m = mass of the body
∆t = Rise in temperature
C = Specific heat capacity of a substance depends on the nature of the material of the substance.
S.I unit of specific heat is J kg-1 K-1.
Given,
An ice cube of mass = m=25g
initial temperature = T1 = -8°C
final temperature = T2 = 0°C
Heat capacity for solid water = c = 2.1J/g°C
According to heat energy required to take it to the melting point,
Q = mc∆T
Q = 25g×2.1J/g°C × (0+8) °C
Q = (25×2.1×8) J
Q = 420J
Hence, the heat energy required to take it to the melting point is 420J.
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PLEASE HELP!!
For this car, the airbag must have a volume of 58 liters when fully inflated. To provide an adequate cushion for the driver’s head, the air pressure inside the airbag should be 4.4 psi. This pressure value is in addition to the normal atmospheric pressure of 14.7 psi, giving a total absolute pressure of 19.1 psi, which equals 1.30 atmospheres.
One of the main components of an airbag is the gas that fills it. As part of the design process, you need to determine the exact amount of nitrogen that should be produced. Calculate the number of moles of nitrogen required to fill the airbag. Show your work. Assume that the nitrogen produced by the chemical reaction is at a temperature of 495°C and that nitrogen gas behaves like an ideal gas.
Part C
Recall the balanced chemical equation from part B of task 1:
2NaN3 → 2Na + 3N2.
Calculate the mass of sodium azide required to decompose and produce the number of moles of nitrogen you calculated in part B of this task. Refer to the periodic table to get the atomic weights.
Part D
What would happen if the amount of sodium azide used was far greater or far less than what you calculated in part C? Describe both cases.
The number of moles of nitrogen required to fill the airbag is 1.197 moles.
The mass of sodium azide required to decompose and produce 1.197 moles of nitrogen is 51.87 g.
If the mass used was greater, the airbag could burst, but if the mass used was smaller, the airbag would not inflate properly.
What amount in moles of nitrogen is required to fill the bag?The number of moles of nitrogen required to fill the airbag is calculated from the equation of reaction as follows:
Using the ideal gas equation; PV = nRT
where;
P = pressureV = volumen = number of molesR = molar gas constantT = temperatureFrom the data provided:
P = 1.30 atm
V = 58 Liters
n = ?
R = 0.082 atm.L.mol⁻¹K⁻¹
T = 495 °C or ( 273.15 + 495) K = 768.15 K
solving for n;
n = PV/RT
n = (1.3 * 58) / (0.082 * 768.15)
n = 1.197 moles
Equation of reaction: 2 NaN₃ → 2 Na + 3 N₂
moles ratio = 2 moles of sodium azide produce 3 moles of N₂
Moles of azide required = 1.197 * 2/3 = 0.798 moles
molar mass of sodium azide = 65 g/mol
mass of sodium azide = 0.798 * 65
mass of sodium azide required = 51.87 g
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Please help with Chemistry I'm confused at the elements reactivity:
Question:
Which of the following combination of elements is the most reactive?
Answer options:
A: Potassium and Iodine
B: Caesium and Iodine
C: Sodium and Bromine
D: Caesium and Bromine
The combination of elements that is the most reactive would be caesium and bromine.
Reactivity of metals and non-metalsThe reactivity series of elements is a table that shows how reactive elements are, usually in descending order.
Elements are classified as metals, non-metals, and metalloids. Metals or non-metals that are at the top of the reactivity series are highly reactive and cannot be displaced in solution by metals below them.
Highly reactive non-metals are also usually at the top of the reactivity series of non-metals. They are able to form compounds with a wide variety of elements.
A typical reactivity series of metals include caesium, rubidium, potassium, sodium, lithium, etc. A reactivity series of non-metals include fluorine, chlorine, oxygen, bromine, iodine, etc.
Thus, bromine is more reactive than iodine and caesium is more reactive than potassium and sodium.
The combination of elements that are the most reactive among the options is, therefore, Caesium and Bromine.
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Your spaceship has docked at a space station above Mars. The temperature inside the space station is a carefully controlled 24 ∘C at a pressure of 745 mmHg . A balloon with a volume of 443 mL drifts into the airlock where the temperature is − 95 ∘C and the pressure is 0.115 atm . What is the final volume, in milliliters, of the balloon if n does not change and the balloon is very elastic?
Answer: the final volume of the balloon is 2.26 x 10^3 mL
Explanation:
The question requires us to determine the new volume of a balloon, given the initial and final conditions.
The following information was provided by the question:
initial temperature = T1 = 24 °C = 297.15 K
initial volume = V1 = 443 mL
initial pressure = P1 = 745 mmHg = 0.980 atm
final temperature = T2 = -95 °C = 178.15 K
final pressure = P2 = 0.115 atm
To solve this problem, we'll need to apply the equation of ideal gases to calculate the number of moles of gas in the balloon, and then use this value and the final temperature and pressure provided to determine the final volume,
The equation of ideal gases can be written as:
[tex]PV=nRT[/tex]And we can rearrange it to calculate the number of moles:
[tex]PV=nRT\rightarrow n=\frac{PV}{RT}[/tex]Applying the values provided by the question:
[tex]n=\frac{(0.980atm)\times(443mL)}{R\times(297.15K)}=\frac{1.46}{R}(\frac{atm\times mL}{K})[/tex]Now, we can rearrange the equation of ideal gases to calculate the volume:
[tex]PV=nRT\rightarrow V=\frac{nRT}{P}[/tex]And, applying the values provided and the number of moles as calculated:
[tex]V=\frac{(\frac{1.46}{R}atm.mL.K^{-1})\times R\times(178.15K)}{0.115atm}=2.26\times10^3mL[/tex]Therefore, the final volume of the balloon is 2.26 x 10^3 mL.
Place the following gases in order of decreasing average molecular speed at 300K: He, SF6, SO2, Ar
Temperature is a way to measure the kinetic energy of the molecules of gases. The kinetic energy is a kind of energy that a moving body has. it is related to the mass of the body and the speed they are moving. The kinetic energy can be calculated as:
[tex]E_k=\frac{1}{2}mv^2[/tex]Where Ek is the kinetic energy, m is mass of the molecules and v is the speed of the molecule
Knowing that gasses with the same temperature have same kinetic energy, then gasses with larger mass will move slower and viceversa.
Since the molar mass of those gases are:
He: 4.00 g/mol
SF6: 146.06 g/mol
SO2: 64.066 g/mol and
Ar: 39.948 g/mol
the ligther gasses move faster and the hevier slower, so we just have to order them by increasing mass:
He, Ar, SO2 and SF6
PLEASE HELP, WILL MARK BRANLIEST!!
The half-life of cobalt-60 is 10.47 min. How many grams of cobalt-60 remain after 104.7 min if you start with 1024g?
Answer: 1 gram
Explanation:
- the equation is 1024(.5)^10
• so 1024g is your initial amount and it’s the amount that is being decayed
• since it’s half-life, the decay factor is 0.5
• the half life is 10.47 minutes, so 10.47 minutes would be considered one round of half life.
-basically, if you were finding the half life after 10.47 minutes you would use the equation 1024(.5)^1.
-as said, it’s raised to the 1st power because 10.47 minutes is one round.
• since you’re looking for the amount after 104.7 minutes, the equation is raised to the 10th power because 104.7 / 10 = 10.47. so 104.7 is 10 rounds of half-life.
- so 1024(.5)^10 = 1
so the half-life of 1024g of Cobalt-60 after 104.7 minutes is 1
hope this helps :)
How many grams of CO2 are produced if 12 liters of H2O are produced?
Answer
[tex]14653.7181\text{ g }CO_2\text{ }[/tex]Procedure
Considering the following balanced equation.
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
To determine the grams of CO₂ that will be produced, you will first need to get the grams of water produced, then convert the grams of water into moles of water and use the molar proportios to get the moles of CO₂. Lastly, convert the moles of CO₂ into grams.
Additional data:
Dnsity of water = 1 g H₂O=/ of H₂O
CO₂ molar mass = 44.01 g/mol
H₂O molar mass = 18.02 g/mol
Calculations
Convert to mass
[tex]12L\text{ }H₂O\text{ }\frac{1kg\text{ }H_2O}{1L\text{ }H_2O}=12kg\text{ }H_2O[/tex]Convert to moles
[tex]12kg\text{ }H₂O\text{ }\frac{1000g\text{ }H_2O\text{ }}{1kg\text{ }H_2O}\frac{\text{ }1\text{ }mole\text{ }H_2O}{18.02g\text{ }H_2O}=665.9267\text{ }mole\text{ }H_2O[/tex]Convert to moles of CO₂ using the relationship
[tex]665.9267\text{ }mole\text{ }H_2O\frac{2moleCO_2}{4mole\text{ }H_2O}=332.9634\text{ }mole\text{ }CO_2[/tex]Convert to grams of CO₂
[tex]332.9634\text{ }mole\text{ }CO_2\text{ }\frac{44.01g\text{ }CO_2}{1\text{ }mole\text{ }CO_2}=14653.7181\text{ g }CO_2\text{ }[/tex]How many moles of argon atoms are present in 11.2 L of argon gas at STP? Round your molar mass to whole numbers andgive your answer in the correct number of sig figs and units
STP or Standard Pressure and Temperature is widely used when the subject is gases. The used standard temperature is 273 K or 0°C and the standard temperature used is 1 atm. At these conditions 1 mol will be equal to 22.4 L of volume, therefore if we have 11.2 L of Argon:
22.4 L = 1 mol
11.2 L = x moles of Argon
22.4x = 11.2 L
x = 0.500 moles of Ar will be present in 11.2 L of volume
What is the transfer of thermal energy between objects?A. HeatB. HotC. TemperatureD. Cold
Thermal energy is transferred from a hot body (higher temperature) to a colder body (lower temperature). In these energy exchanges, two types of changes are observed: temperature or physical state. These variations occur both with energy intake and loss. What is being transferred between the two bodies is heat.
Answer: A. Heat.
Which type of element is shown in green boxes
Planets A, B, and C revolve around stars like the sun in orbits like that of Earth. Could planet A possibly support human life?
The planet, known as Proxima b, may be warm enough for liquid water to exist on its surface and hence be suitable for life because it is located within the star's "habitable zone."
The first and second laws of Kepler were what?Applying Kepler's laws: The sun is at the centre of elliptical planetary orbits, according to the first law. Second Law: The radius vector from the sun to a planet covers the same area in exactly the same amount of time. Third Law: For every planet, there is a constant relationship between the cube of the elliptical semimajor axis and the square of the period of revolution.
Since the angular momentum is changing at a rate of zero, the angular momentum must be constant, which implies that the rate of change of the swept-out area for the celestial body's orbit must also be constant. This leads to Kepler's Second Law, which states that celestial objects in orbit cover equal regions in similar amounts of time.
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. At standard pressure, what state of matter is xenon at -111 °C?
A) solid
B) liquid
C) gas
D) can be both solid and liquid
E) can be both liquid and gas
At standard pressure, C) gas state of matter is xenon at -111 °C
What is the state of matter of xenon?Xenon is an extremely uncommon, odourless, colourless, tasteless, chemically inert gas. It was thought to be absolutely innocuous until Neil Bartlett published the synthesis of xenon haxafluoroplatinate in 1962. When stimulated by an electrical discharge, xenon generates blue light in a gas-filled tube.
Noble gases, which are most commonly encountered as monatomic gases, have entirely filled outer electron shells and hence have little desire to react with other elements, resulting in relatively few compounds with other elements.
In its +6 oxidation state, xenon trioxide is an unstable molecule. It is a highly potent oxidising agent that progressively liberates oxygen from water when exposed to sunshine. When it comes into touch with organic materials, it becomes highly explosive.
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help me please if you can a. ammonia b. battery acid c.pure waterd. sea water
Answer:
ammonia
Explanation:
How many atoms of oxygen are in the reactants of the following equation? 3NO2 + H2O --> NO + 2HNO3
The reactants are the compounds before the arrow and the products are after the arrow. Just as they show us, the oxygen atoms in the reagents will bJust as the reaction shows, the oxygen atoms in the reactants will be:
Answer: There are 7 atoms of oxygen in the reactants
If 1495 J of heat is needed to raise the temperature of a 315 g sample of a metal from 55.0°C to 66.0°C, what is the specific heat capacity of the metal?
The specific heat capacity of the metal that needs 1495 J of heat to raise the temperature is 0.43J/g°C.
How to calculate specific heat capacity?Specific heat capacity is the heat capacity per unit mass of a substance. It can be calculated by using the following formula:
Q = mc∆T
Where;
Q = quantity of heat absorbed or releasedm = mass of substancec = specific heat capacity∆T = change in temperatureAccording to this question, 1495 J of heat is needed to raise the temperature of a 315 g sample of a metal from 55.0°C to 66.0°C. The specific heat capacity can be calculated as follows;
1495 = 315 × c × {66 - 55}
1495 = 3465c
c = 1495/3465
c = 0.43J/g°C
Therefore, 0.43J/g°C is the specific heat capacity of the metal that requires a heat of 1495J.
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when a bunsen burner is properly adjusted, what should the flame look like?
when a bunsen burner is properly adjusted, the flame looks Blue with no gap between the burner and the flame.
A Bunsen burner is a particular kind of gas burner that is frequently utilized as a heat source in lab investigations. The barrel or chimney of the burner is made up of a straight tube that extends vertically from a flat base. At the bottom of the chimney, liquid petroleum gas, such as propane or butane, or natural gas, primarily methane, is delivered.
The base of the chimney on Bunsen burners typically has a hose barb installed so that rubber tubing may supply the gas from a gas nozzle on the lab bench.
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A compound containing 63.88% Cl and 36.12% Ca has a molecular mass of 443.92 g/mol, what is the molecular formula?
The molecular formula of a compound containing 63.88% Cl and 36.12% Ca that has a molecular mass of 443.92 g/mol is Ca₄Cl₈.
How to calculate molecular formula?Molecular formula is the notation indicating the number of atoms of each element present in a compound.
To calculate the molecular formula of a compound, the empirical formula must first be calculated as follows:
63.88% Cl = 63.88g ÷ 35.5g/mol = 1.79mol36.12% Ca = 36.12g ÷ 40g/mol = 0.903molCl = 1.79mol ÷ 0.903 = 1.98Ca = 0.903mol ÷ 0.903 = 1The approximate empirical ratio of Cl and Ca is 2:1, hence, the empirical formula is CaCl₂.
{CaCl₂}n = 443.92
{(40 + 35.5(2)}n = 443.92
111n = 443.92
n = 4
The molecular formula is Ca₄Cl₈.
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Answer:
Ca4Cl8
Explanation:
Canva
2022
a) A solution of sodium thiosulfate, Na2S2O3, is 0.1047 M. 18.59 mL of this solution reacts with 38.62 mL of I2 solution. What is the molarity of the I2 solution?
2(Na2S2O3) + I2↔Na2S4O6 + 2(NaI)
b) 26.64 mL of the I2 solution from above is required to titrate a sample containing As2O3. Calculate the mass of As2O3 (197.8 g/mol) in the sample.
As2O3 + 5(H2O) + 2I2 → 2(H3AsO4) + 4HI
From the calculations, the molarity of the iodine solution is 0.025 M and the mass of the arsenic oxide solution is 0.066 g.
What is the molarity?Let us recall that the molarity is defined as the ratio of the number of moles to the volume of the solution. It is a measure of the amount of substance present. Ley us now try to use what we know to obtain the molarity of the solution in each of the cases of the questions.
a) Using;
CAVA/CBVB = NA/NB
CA = Amount of thiosuphate
CB = Amount of iodine
VA = volume of thiosuphate
VB = volume of iodine
NA = number of moles of thisulphate
NB = Number of moles of iodine
Hence;
0.1047 M * 18.59 mL/CB * 38.62 mL = 2/1
0.1047 M * 18.59 mL * 1 = CB * 38.62 mL * 2
CB = 0.1047 M * 18.59 mL * 1 /38.62 mL * 2
CB = 1.946/77.2
CB = 0.025 M
b) Using the formula;
Number of moles = concentration * volume
Number of moles = 0.025 M * 26.64 /1000 L
= 6.66 * 10^-4 moles
Again;
1 mole of the arsenic oxide reacts with 2 moles of iodine solution
x moles of the arsenic oxide reacts with 6.66 * 10^-4 moles of iodine
x = 1 mole * 6.66 * 10^-4 moles/ 2 moles
x = 3.33 * 10^-4 moles
Mass of the arsenic oxide = 3.33 * 10^-4 moles * 197.8 g/mol
= 0.066 g
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What is the concentration in molarity of a solution which is 2.91 %m/v benzene (C₆H₆, MM =78.11 g/mol ) in CCl₄ (MM = 153.81 g/mol)?
According to molar concentration,concentration in molarity of the solution is 3.77×10[tex]^-7[/tex] M.
What is molar concentration?Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.
The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.
In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.
In the given problem,mass of benzene =2.91 g and molar mass of benzene=78.11 g/mole and volume=100-2.91=97.09 ml or 97.09×10[tex]^-3[/tex] L.
Substituting in the formula,molarity=2.91/78.11×1/97.09×10[tex]^-3[/tex]=3.77×10[tex]^-7[/tex] M.
Hence, the molarity of solution is 3.77×10[tex]^-7[/tex] M.
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metal oxide MO2 reacts with excess HCl to produce chlorine gas at STP as given by the following unbalanced equation:MO2 (s) + HCl (aq) -,> MCl2 (aq) + Cl2 (g) + H20 (l) a. determine limiting reactantb. calculate the mass of MCl2 produced in the reactionc. calculate the percentage yield if the actual mass of MCl2 produced is 0.078g
1st) It is necessary to balance the chemical equation:
[tex]MO_2+4\text{HCl}\rightarrow MCl_2+Cl_2+2H_2O[/tex]2nd)
DUE AT 11:59 PLEASE HELP
A student has a calorimeter with 211.7 grams of 20.4 degrees Celsius water contained within it. The student then adds 128.9 grams of 94.2 degrees Celsius water to that calorimeter and stirs. To what maximum temperature will the cold water in the calorimeter rise to?
The maximum temperature the cold water in the calorimeter will rise to is 48.3 °C
How to determine the maximum temperatureThe maximum temperature the cold water can attain can be obtained by calculating the equilibrium temperature. This can be obtained as follow:
From the question given above, the following data were obtained:
Mass of cold water (M) = 211.7 grams Temperature of cold water (T) = 20.4 °CMass of warm water (Mᵥᵥ) = 128.9 gramsTemperature of warm water (Tᵥᵥ) = 94.2 °CSpecific heat capacity of the water (C) = 4.18 J/gºC Equilibrium temperature (Tₑ) =?Heat loss = Heat gain
MᵥᵥC(Tᵥᵥ – Tₑ) = MC(Tₑ – T)
Cancel out C
Mᵥᵥ(Tᵥᵥ – Tₑ) = M(Tₑ – T)
128.9 × (94.2 – Tₑ) = 211.7 × (Tₑ – 20.4)
Clear bracket
12142.38 – 128.9Tₑ = 211.7Tₑ – 4318.68
Collect like terms
12142.38 + 4318.68 = 211.7Tₑ + 128.9Tₑ
16461.06 = 340.6Tₑ
Divide both side by 340.6
Tₑ = 16461.06 / 340.6
Tₑ = 48.3 °C
Thus, we can conclude that the maximum temperature is 48.3 °C
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This type of bond is between two metals described as a "Sea of Electrons'?
Answer:
In the early 1900's, Paul Drüde came up with the "sea of electrons" metallic bonding
I need help on balancing the equations and on what type of reaction it is.
Given the balanced chemical reaction expressed as:
[tex]2Au_2O_3\rightarrow4Au+3O_2[/tex]A Redox reaction is a reaction that involves the transfer of electrons and changes in the oxidation state between the elements.
The given chemical equation is therefore an oxidation-reduction reaction since it involves a change in the oxidation state of the elements.
Determine the moles of Au₂O₃
[tex]\begin{gathered} \text{Mole = }\frac{Mass}{Molar\text{ mass}} \\ \text{Mole of Au}_2O_3=\frac{10g}{441.93g\text{/mol}} \\ \text{Mole of Au}_2O_3=0.02263\text{moles} \end{gathered}[/tex]According to stochiometry, you can see that 2 moles of Gold(III)oxide produce 4 moles of Gold. Hence the moles of Gold produced will be:
[tex]\begin{gathered} \text{moles of Gold=}\frac{0.02263\times4}{2} \\ \text{moles of Gold=}0.02263\times2 \\ \text{moles of Gold=}0.0453\text{moles} \end{gathered}[/tex]Determine the mass of Gold.
[tex]\begin{gathered} \text{Mass of Gold=moles}\times molar\text{ mass} \\ \text{Mass of Gold=}0.0453\times196.97 \\ \text{Mass of Gold}=8.91\text{grams} \end{gathered}[/tex]Next is determining the mole of Oxygen
According to stochiometry, you can see that 2 moles of Gold(III)oxide produce 3 moles of Oxygen. Hence the moles of Oxygen produced will be:
[tex]\begin{gathered} \text{moles of Oxygen=}\frac{0.02263\times3}{2} \\ \text{moles of Oxygen}=0.033945\text{moles} \end{gathered}[/tex]Determine the mass of oxygen produced
[tex]\begin{gathered} \text{Mass of O}_2=moles\times\text{Molar mass} \\ \text{Mass of O}_2=0.033945\times16 \\ \text{Mass of O}_2=0.543\text{grams} \end{gathered}[/tex]Hence the mass of oxygen produced is 0.543 grams
what is the limiting and excess reactant if 15.0g of FePo4 reacts with 5.0g of Na2SO
Explanation:
We are given: mass of FePO4 = 15g
: mass of Na2SO4 = 5g
We first find the mass of Fe2(SO4)3 from the mass of FePO4:
m is the mass and M is the molar mass
[tex]\begin{gathered} m\text{ = }\frac{m(FePO4)}{M(FePO4)}\times\frac{1mol\text{ Fe2\lparen SO4\rparen3}}{2mol\text{ FePO4}}\times\text{ M\lparen Fe2\lparen SO4\rparen3\rparen} \\ \text{ = }\frac{15}{150.82}\times\frac{1}{2}\times399.88 \\ \text{ = 19.89g} \end{gathered}[/tex]We then find the mass of Fe2(SO4)3 from Na2SO4:
[tex]\begin{gathered} m\text{ = }\frac{m(Na2SO4)}{M(Na2SO4)}\times\frac{1mol\text{ Fe2\lparen SO4\rparen3}}{3mol\text{ Na2SO4}}\times M(Fe2(SO4)3) \\ \text{ =}\frac{5}{142.04}\times\frac{1}{3}\times399.88 \\ \text{ = 4.69g} \end{gathered}[/tex]Answer:
Therefore, FePO4 is the excess reactant and Na2SO4 is the limiting reactant
an equimolar mixture of n2(g)n2(g) and ar(g)ar(g) is kept inside a rigid container at a constant temperature of 300 kk. the initial partial pressure of arar in the mixture is 0.75atm0.75atm. an additional amount of arar was added to the container, enough to double the number of moles of arar gas in the mixture. assuming ideal behavior, what is the final pressure of the gas mixture after the addition of the arar gas? responses 0.75atm0.75atm, because increasing the partial pressure of arar decreases the partial pressure of n2n2. 0.75 atmosphere , because increasing the partial pressure of a r decreases the partial pressure of n 2 . 1.13atm1.13atm, because 333% of the moles of gas are n2n2. 1.13 atmospheres , because 33 percent of the moles of gas are n 2 . 1.50atm1.50atm, because the number of moles of n2n2 did not change. 1.50 atmospheres , because the number of moles of n 2 did not change. 2.25atm2.25atm, because doubling the number of moles of arar doubles its partial pressure.
To solve such type of question we must be knowing the concept behind the ideal gas equation. The final pressure of the gas mixture after the addition of the Ar gas is 2.25 atm
What is ideal gas equation?Ideal gas equation is the mathematical expression that relates pressure volume, temperature and number of moles of gas
Mathematically,
PV=nRT
according to question T and V is constant
P ∝ n
P₁/n₁= P₂/n₂
Where
P₁ = initial pressure= 0.75atm
P₂ = Final pressure=?
n₁= number of moles of gas initially present=n
n₂ = Final moles of gas present=2n
Substituting into the given equation
P₂= P₁n₂/n₁
P₂ = 0.75atm ×2n/n
P₂ = 1.5 atm
The total pressure of the gas=partial pressure of N2 + partial pressure after addition of Ar = 0.75 atm + 1.5 atm = 2.25 atm
The final pressure of the gas mixture after the addition of the Ar gas is 2.25 atm
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Calculate the pH of the resulting solution when 25.0mL of 0.30M HClO4 is added to 60.0ml of 0.35 M CH3NH2. Kb for CH3NH2 = 4.4*10^-4
The pH of the resulting solution is 10.9 .
given that :
for HClO₄
volume = 25 mL = 0.025 L
M = 0.30 M
no. of moles = 0.025 × 0.30
= 0.0075
for CH₃NH₂
no. of moles = 0.35 × 0.06
= 0.021
as a resultant no. of moles of CH₃NH₂ that remain in solution
= 0.021-0.0075
= 0.0135
total volume = 0.025 + 0.060
= 0.085
the concentration will be :
concentration for HClO₄ = 0.0075 / 0.085
= 0.088 M
concentration for CH₃NH₂ = 0.0135 / 0.088
= 0.153
kb = 4.4 × 10⁻⁴
pkb = - log kb
pkb = 3.3
the pOH formula is given as :
POH = pKb + log [ acid ] / [ base ]
pOH = 3.3 + log [ 0.088 ] / [ 0.153 ]
pOH = 3.06
pH + pOH = 14
pH = 14 - 3.06
pH = 10.9
Thus, The pH of the resulting solution when 25.0mL of 0.30M HClO₄ is added to 60.0ml of 0.35 M CH₃NH₂. Kb forCH₃NH₂ = 4.4 × 10^-4 . pH is 10.9.
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Calculate the imperial formula of the compound. Express your answer as a chemical formula.
The empirical formula corresponds to the simplest form of expressing a compound, it indicates the proportion of atoms in the molecule.
We have two elements S and F and they give us the mass resulting from the decomposition of the molecule. We can find the moles of each element using the atomic weight of each element as follows:
[tex]\begin{gathered} \text{Moles S= Given g of S}\times\frac{1\text{ mol S}}{AtomicWeight,\text{ g S}} \\ \text{Moles S= 0.905 g of S}\times\frac{1\text{ mol S}}{32.065\text{ g S}}=0.028\text{mol S} \end{gathered}[/tex][tex]\begin{gathered} \text{Moles F= Given g of F}\times\frac{1\text{ mol F}}{AtomicWeight,\text{ g F}} \\ \text{Moles F=3.221 g of F}\times\frac{1\text{ mol F}}{18.998\text{g F}}=0.170\text{mol F} \end{gathered}[/tex]To find the ratio between the elements we divide the moles of each element by the smallest number of moles found, that is by 0.028 moles.
[tex]\begin{gathered} S\rightarrow\frac{0.028}{0.028}=1 \\ F\rightarrow\frac{0.170}{0.028}=6 \end{gathered}[/tex]Therefore the empirical formula of the compound will be:
[tex]SF_6[/tex]If you start with 29.0 grams of Ca(OH)2 and 12.5 grams of HCI how many moles of water can be formed?
Answer
The moles of water that will be produced = 0.343 mol
Explanation
Given:
Mass of Ca(OH)2 = 29.0 g
Mass of HCl = 12.5 g
We know the reaction : Ca(OH)2 + 2HCI —> CaCI2 + 2H2O
Molar mass of Ca(OH)2 = 74.093 g/mol
Molar mass of HCl = 36.458 g/mol
Molar mass of H2O = 18.015 g/mol
Required: Moles of water that will be formed
Solution:
Use the stoichiometry to find the moles of water using both the reactants.
For Ca(OH)2:
[tex]\begin{gathered} 29.0\text{ g Ca\lparen OH\rparen}_2\text{ x }\frac{1\text{ mole Ca\lparen OH\rparen}_2}{74.093\text{ g Ca\lparen OH\rparen}_2}\text{ x }\frac{2\text{ mol H}_2O}{1\text{ mol Ca\lparen OH\rparen}_2} \\ \\ =\text{ 0.783 mol H}_2O \end{gathered}[/tex]For HCl
[tex]\begin{gathered} 12.5\text{ g HCl x }\frac{1\text{ mol HCl}}{36.458\text{ g}}\text{ x }\frac{2\text{ mol H}_2O}{2\text{ mol HCl}} \\ \\ =\text{ 0.343 mol H}_2O \end{gathered}[/tex]HCl will produce less moles of H2O, thus HCl is the limiting reactant/reagent and the moles of water that will be produced = 0.343 mol
I need to understand how to do problem number 5
Answer:
[tex]\begin{gathered} \text{ - molecule of H}_2SO_4=2.408\times10^{24} \\ -4\text{ atoms of Oxygen} \end{gathered}[/tex]Explanations:
Given the following parameters
Moles of sulfuric acid = 14 moles
According to the Avogadro's constant;
[tex]1\text{mole}=6.02\times10^{23}molecules[/tex]The number of molecules of 14 moles of sulfuric acid is calculated as:
[tex]\begin{gathered} \text{ molecule of H}_2SO_4=14\times6.02\times10^{23} \\ \text{ molecule of H}_2SO_4=24.08\times10^{23} \\ \text{ molecule of H}_2SO_4=2.408\times10^{24} \end{gathered}[/tex]Hence the molecule of sulfuric acid that is contained in 14moles if sulfuric acid is 2.408 * 10^24 molecules
Since the chemical formula of sulfuric acid is expressed as H₂SO₄. This shows that the compound has 2 atoms of Hydrogen, 1 atom of sulfur, and 4 atoms of oxygen.
Hence the number of atoms of oxygen contained in this sample is 4 atoms