Question

1. nitrogen gas was collected over water at 25°c. if the vapor pressure of water at 25°c is 23 mmhg, and the total pressure in the container is 781 mmhg, what is the partial pressure of the nitrogen gas?
2. a 22.0 g sample of an unknown gas occupies 11.2 l at stp. what is the identity of the gas from these 4 choices? co₂, so₃, o₂ or he?
3. gas molecule a weighs 4 times less than gas molecule b. how much faster does gas molecule a have to move in comparison to gas molecule b for them to have the same kinetic energy?

Explanation:

Question 11

Step1: Recall Dalton's law of partial pressures

The total pressure ($P_{total}$) of a gas - mixture is the sum of the partial pressures of its components. When nitrogen gas is collected over water, $P_{total}=P_{N_2}+P_{H_2O}$. We want to find $P_{N_2}$.

Step2: Rearrange the formula

$P_{N_2}=P_{total}-P_{H_2O}$

Step3: Substitute the given values

Given $P_{total} = 781\ mmHg$ and $P_{H_2O}=23\ mmHg$. So, $P_{N_2}=781 - 23$
$P_{N_2}=758\ mmHg$

Step1: Recall the molar volume at STP

At STP (Standard Temperature and Pressure, $T = 273\ K$ and $P = 1\ atm$), the molar volume of any ideal gas is $V_m=22.4\ L/mol$.

Step2: Calculate the number of moles of the unknown gas

We know that $n=\frac{V}{V_m}$. Given $V = 11.2\ L$ and $V_m = 22.4\ L/mol$, so $n=\frac{11.2\ L}{22.4\ L/mol}=0.5\ mol$

Step3: Calculate the molar mass of the unknown gas

The molar mass $M=\frac{m}{n}$. Given $m = 22.0\ g$ and $n = 0.5\ mol$, so $M=\frac{22.0\ g}{0.5\ mol}=44\ g/mol$

Step4: Identify the gas

The molar mass of $CO_2$ is $M_{CO_2}=12 + 2\times16=44\ g/mol$, the molar mass of $SO_3$ is $M_{SO_3}=32+3\times16 = 80\ g/mol$, the molar mass of $O_2$ is $M_{O_2}=2\times16 = 32\ g/mol$, and the molar mass of $He$ is $M_{He}=4\ g/mol$. So the gas is $CO_2$.

Step1: Recall the kinetic - energy formula

The kinetic energy of a gas molecule is $K.E.=\frac{1}{2}mv^{2}$. Let the mass of gas molecule B be $m_B$ and its speed be $v_B$, and the mass of gas molecule A be $m_A$ and its speed be $v_A$. We know that $m_A=\frac{1}{4}m_B$. Since $K.E._A = K.E._B$, we have $\frac{1}{2}m_Av_A^{2}=\frac{1}{2}m_Bv_B^{2}$

Step2: Substitute $m_A=\frac{1}{4}m_B$ into the kinetic - energy equation

$\frac{1}{2}\times\frac{1}{4}m_Bv_A^{2}=\frac{1}{2}m_Bv_B^{2}$

Step3: Solve for the ratio of the speeds

Cancel out $\frac{1}{2}m_B$ from both sides of the equation, we get $\frac{1}{4}v_A^{2}=v_B^{2}$. Then $v_A^{2}=4v_B^{2}$, and $v_A = 2v_B$

Answer:

$758\ mmHg$

Question 12