1. What is the volume in liters of 8.20 moles of CO₂ at standard temperature and pressure (STP)?2. A sample of 3.05 mol of gas in a 10.00 L container is at 45.0 °C. What is the pressure (in atm) of the gas?
3. What temperature (in °C) did an ideal gas shift to if it was initially at -17.00 °C at 4.620 atm and 35.00 L and the pressure was changed to 8.710 atm and the volume changed to 15.00 L?
4. A mixture of two gases with a total pressure of 1.98 atm contains 0.70 atm of Gas A. What is the partial pressure of Gas B in atm?
5. A chamber contains equal molar amounts of He, Ne, Ar, and Kr. If the total chamber pressure is 1 atm, then the partial pressure (in atm) of Kr is:

Answers

Answer 1

Answer:

Answer:

1. To find the volume of 8.20 moles of CO₂ at standard temperature and pressure (STP), we can use the ideal gas law. At STP, the temperature is 0 °C or 273.15 K, and the pressure is 1 atm. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

We can rearrange the equation to solve for V:

V = (nRT) / P

Substituting the values:

V = (8.20 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

Calculating the numerical value:

V ≈ 181.3 L

Therefore, the volume of 8.20 moles of CO₂ at STP is approximately 181.3 liters.

2. To find the pressure of a gas sample with 3.05 moles in a 10.00 L container at 45.0 °C, we can still use the ideal gas law. However, we need to convert the temperature to Kelvin by adding 273.15 to it.

The ideal gas law equation can be rearranged to solve for pressure:

P = (nRT) / V

Substituting the values:

P = (3.05 mol * 0.0821 L·atm/mol·K * (45.0 + 273.15) K) / 10.00 L

Calculating the numerical value:

P ≈ 4.083 atm

Therefore, the pressure of the gas sample is approximately 4.083 atm.

3. To find the final temperature in °C when the initial temperature was -17.00 °C, and the pressure changed from 4.620 atm to 8.710 atm, and the volume changed from 35.00 L to 15.00 L, we can use the combined gas law.

The combined gas law states that (P₁ V₁) / T₁ = (P₂ V₂) / T₂, where P is pressure, V is volume, and T is temperature.

Rearranging the equation to solve for T₂:

T₂ = (P₂ V₂ T₁) / (P₁ * V

Answer 2

Answer:

Final answer:

The volume of 8.20 moles of CO₂ at standard temperature and pressure (STP) is approximately 180.4 liters.

Explanation:

Gas Laws

Gas laws describe the behavior of gases under different conditions. One of the fundamental gas laws is the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Question 1: Volume of CO₂ at STP

To find the volume of 8.20 moles of CO₂ at standard temperature and pressure (STP), we can use the ideal gas law equation. At STP, the temperature is 0 degrees Celsius (273.15 Kelvin) and the pressure is 1 atmosphere (atm).

Given:

Number of moles (n) = 8.20 moles
Temperature (T) = 0 degrees Celsius (273.15 Kelvin)
Pressure (P) = 1 atmosphere (atm)

Using the ideal gas law equation, we can rearrange it to solve for the volume (V):

V = (nRT) / P

Substituting the given values:

V = (8.20 moles * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

V ≈ 180.4 liters

Question 2: Pressure of Gas

To find the pressure of a gas in a given container, we can use the ideal gas law equation. The temperature must be converted to Kelvin before using the equation.

Given:

Number of moles (n) = 3.05 mol
Volume (V) = 10.00 L
Temperature (T) = 45.0 °C (318.15 Kelvin)

Using the ideal gas law equation, we can rearrange it to solve for the pressure (P):

P = (nRT) / V

Substituting the given values:

P = (3.05 mol * 0.0821 L·atm/mol·K * 318.15 K) / 10.00 L

P ≈ 7.79 atm

Question 3: Temperature Change

To find the temperature change of an ideal gas, we can use the ideal gas law equation. The initial and final conditions of the gas must be known.

Given:

Initial temperature (T1) = -17.00 °C (256.15 Kelvin)
Initial pressure (P1) = 4.620 atm
Initial volume (V1) = 35.00 L
Final pressure (P2) = 8.710 atm
Final volume (V2) = 15.00 L

Using the ideal gas law equation, we can rearrange it to solve for the final temperature (T2):

T2 = (P2 * V2 * T1) / (P1 * V1)

Substituting the given values:

T2 = (8.710 atm * 15.00 L * 256.15 K) / (4.620 atm * 35.00 L)

T2 ≈ 303.6 °C

Question 4: Partial Pressure of Gas B

To find the partial pressure of Gas B in a mixture of gases, we need to know the total pressure and the partial pressure of Gas A.

Given:

Total pressure = 1.98 atm
Partial pressure of Gas A = 0.70 atm

The partial pressure of Gas B can be calculated by subtracting the partial pressure of Gas A from the total pressure:

Partial pressure of Gas B = Total pressure - Partial pressure of Gas A

Partial pressure of Gas B = 1.98 atm - 0.70 atm

Partial pressure of Gas B ≈ 1.28 atm

Question 5: Partial Pressure of Kr

To find the partial pressure of Kr in a chamber containing equal molar amounts of He, Ne, Ar, and Kr, we need to know the total chamber pressure.

Given:

Total chamber pressure = 1 atm

Since the chamber contains equal molar amounts of gases, the partial pressure of Kr is equal to the total chamber pressure divided by the number of gases:

Partial pressure of Kr = Total chamber pressure / Number of gases

Partial pressure of Kr = 1 atm / 4

Partial pressure of Kr = 0.25 atm

Learn more about gas laws here:

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Answers

Answer:

b acid have ph less than 7

describe the process that ice on mount everest goes through when being heated from 10 degree fahrenheit to 305 degree kelvin. then explain how the energy requiresment change at different steps of the heating curve. and what are the energy transfers that happen in the ice on mount everest everest?

Answers

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A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of a 2.18 g sample burns in excess oxygen yields 3.94 g of CO2 and 1.89 g of H2O. A separate experiment shows that a 1.23 g sample contains 0.235 g of N. Calculate the moles of C in the sample.1. Calculate the moles of N for a 2.18 g sample (the sample size for the other analyses).
2. Calculate the moles of O in the sample.

Answers

Answer:

The answers are

C = 0.0895 moles

1. 0.02975 moles of N

2. 0.0299 moles O

Explanation:

Mass of sample = 2.18 g

Combustion yields 3.94 g of CO₂ and 1.89 g of H₂O

1.23 g of sample contains 0.235 g of N

3.94 g of CO₂ which contains one mole of C and two moles of O

Molar mass of CO₂ = 44 mass fraction of C = 12/44 = 0.273, Therefore mass of carbon in 2.18 g = 0.273×3.94 = 1.075 g

number of moles of C in sample = mass/(molar mass) = 1.075 g/(12.01 g/mol) = 0.0895 moles of C

(1) Mass of nitrogen, in 1.23 g of sample = 0.235

therefore mass in 2.18 g =0.235×2.18/1.23 = 0.4165 g

= 0.02975 moles of N

(2) To find the number of moles of oxygen in the sample we note that total mass of sample = 2.18 g containing only nitrogen, hydrogen, oxygen and carbon

Mass of hydrogen can be fround from 2/18×1.89 = 0.21 g

Mass of oxygen therefore = 2.18 -(0.21+0.4165+1.075) = 0.4785

number of moles = 0.475/16 = 0.0299 moles

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Answers

Answer:

Explanation:

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A and D mean the same thing but i believe the answer your looking for is D

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Answers

When the two compounds are allowed to be in contact and reacts the product would be sodium nitrate (NaNO3) and calcium carbonate (CaCO3). This reaction is a double displacement reaction. The balanced equation is :

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50 grams of acetic acid C2H4O2 are dissolved in 200 g of water. Calculate the weight % and mole fraction of the acetic acid in the solution

Answers

$50g$ $of$ $C_2H_4O_2$ $in$ $200g$ $H_2O$

so:

$(50)/(200+50)+100\%= (50*100)/(250)\%=20\%$

$MW of acid = 2*C+4*H+2*O = 2*12+4*1+2*16=$

$=24+4+32=60g/mol$

so:

$(50g)/(60g/mol)\approx0.83moles$

it means that

in $50g$ of acid there is $\approx0.83$ moles of acid

$MW of H_2O = 2*H+O=2*1+16=2+16=18g/mol$

so:

$(200g)/(18g/mol)\approx11.11moles$

it means that:

in $200g$ of water there is $\approx11.11$ moles of water

therefore:

$(0.83mol)/(11.11mol+0.83mol)= (0.83mol)/(11.94mol)=0.069$

So your answers are:

$20\%$

and the mole fraction is:

$0.069$

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