11a. when a sample of zinc reacted with excess hydrochloric acid, 340. ml of hydrogen were collected over water at 30.°c. if the atmospheric pressure was 0.963 atm and the vapor pressure of water at 30.°c is 32 torr, find the mass of zinc that was consumed.
11b. potassium chlorate decomposes into potassium chloride and oxygen upon heating. a 0.7811 - g mixture contains an unknown amount of potassium chlorate. when the mixture is heated, 47.2 ml of oxygen are collected over water at 22°c and 714 torr. the vapor pressure of water at 22°c is 19.8 torr. what mass percent of the mixture was potassium chlorate?
12a. assuming uniform temperature, do all atoms in a given sample of neon have the same kinetic energy?
12b. does a 1 - mol sample of ar have the same total ke as a 1 - mol sample of rn, at a given temperature?
13a. calculate the rms speed of nitrogen dioxide at 85°c.

Question

Accepted Answer

### 13A. Calculate the rms speed of nitrogen dioxide at 85°C.

# Explanation:
## Step 1: Recall the formula for rms speed
The root - mean - square (rms) speed of a gas molecule is given by the formula $v_{rms}=\sqrt{\frac{3RT}{M}}$, where $R$ is the gas constant ($R = 8.314\space J/(mol\cdot K)$ or $R=0.0821\space L\cdot atm/(mol\cdot K)$, but for SI units of speed (m/s), we use $R = 8.314\space J/(mol\cdot K)$), $T$ is the absolute temperature in Kelvin, and $M$ is the molar mass of the gas in $kg/mol$.

## Step 2: Convert temperature to Kelvin
The temperature $T = 85^{\circ}C+ 273.15=358.15\space K$

## Step 3: Determine the molar mass of $NO_2$
The molar mass of $N$ is $14.01\space g/mol$ and the molar mass of $O$ is $16.00\space g/mol$. For $NO_2$, $M=(14.01 + 2\times16.00)\space g/mol=46.01\space g/mol = 0.04601\space kg/mol$

## Step 4: Substitute values into the formula
Substitute $R = 8.314\space J/(mol\cdot K)$, $T = 358.15\space K$ and $M = 0.04601\space kg/mol$ into the formula $v_{rms}=\sqrt{\frac{3RT}{M}}$

$$
\begin{align*}
v_{rms}&=\sqrt{\frac{3\times8.314\times358.15}{0.04601}}\
&=\sqrt{\frac{3\times8.314\times358.15}{0.04601}}\
&=\sqrt{\frac{8930.44}{0.04601}}\
&=\sqrt{194097.8}\
&\approx440.6\space m/s
\end{align*}
$$

# Answer:
The rms speed of nitrogen dioxide at $85^{\circ}C$ is approximately $\boldsymbol{441\space m/s}$ (rounded to three significant figures)

### 12B. Does a 1 - mol sample of Ar have the same total KE as a 1 - mol sample of Rn, at a given temperature?

# Brief Explanations:
The average kinetic energy of a gas molecule is given by $KE_{avg}=\frac{3}{2}kT$ (where $k$ is Boltzmann's constant) and the total kinetic energy of a gas sample is $KE_{total}=n\times\frac{3}{2}RT$ (where $n$ is the number of moles, $R$ is the gas constant, and $T$ is the absolute temperature). For a 1 - mol sample ($n = 1$) of both Ar and Rn at the same temperature $T$, the total kinetic energy depends only on the number of moles and the temperature. Since $n = 1$ for both and $T$ is the same, from the formula $KE_{total}=\frac{3}{2}nRT$, the total kinetic energy of 1 - mol of Ar and 1 - mol of Rn at the same temperature will be the same.

# Answer:
Yes, a 1 - mol sample of Ar has the same total KE as a 1 - mol sample of Rn at a given temperature.

### 12A. Assuming uniform temperature, do all atoms in a given sample of neon have the same kinetic energy?

# Brief Explanations:
The kinetic energy of a gas molecule is given by $KE=\frac{1}{2}mv^{2}$. In a gas sample at a uniform temperature, the average kinetic energy of the gas molecules is given by $KE_{avg}=\frac{3}{2}kT$ (where $k$ is Boltzmann's constant and $T$ is the absolute temperature). However, individual gas molecules have different speeds (due to collisions and the Maxwell - Boltzmann distribution of speeds). Since $KE=\frac{1}{2}mv^{2}$ and the speeds $v$ of individual neon atoms are not the same (they follow a distribution), the kinetic energy of individual neon atoms is not the same. But the average kinetic energy of all the neon atoms is the same at a given temperature.

# Answer:
No, individual atoms in a sample of neon do not have the same kinetic energy. However, the average kinetic energy of the neon atoms is the same at a given temperature.

### 11B. Potassium chlorate decomposes into potassium chloride and oxygen upon heating... (To solve this, we need to follow these steps)

# Explanation:
## Step 1: Write the balanced chemical equation
The decomposition reaction of potassium chlorate ($KClO_3$) is $2KClO_3(s)\xrightarrow{\Delta}2KCl(s)+3O_2(g)$

## Step 2: Find the partial pressure of $O_2$
The total pressure of the gas mixture $P_{total}=714\space torr$, and the vapor pressure of water $P_{H_2O}=19.8\space torr$ at $22^{\circ}C$. By Dalton's law of partial pressures, $P_{O_2}=P_{total}-P_{H_2O}$
$P_{O_2}=714 - 19.8=694.2\space torr$. Convert this pressure to atm: $P_{O_2}=\frac{694.2}{760}\space atm\approx0.9134\space atm$

## Step 3: Find the moles of $O_2$ using the ideal gas law
The volume of $O_2$ collected $V = 47.2\space mL=0.0472\space L$, temperature $T = 22^{\circ}C+273.15 = 295.15\space K$, $R = 0.0821\space L\cdot atm/(mol\cdot K)$
From the ideal gas law $PV = nRT$, we can solve for $n$ (moles of $O_2$):
$n_{O_2}=\frac{P_{O_2}V}{RT}=\frac{0.9134\times0.0472}{0.0821\times295.15}$
$n_{O_2}=\frac{0.0431}{24.23}\approx0.00178\space mol$

## Step 4: Use stoichiometry to find moles of $KClO_3$
From the balanced equation, the mole ratio of $KClO_3$ to $O_2$ is $2:3$. Let $n_{KClO_3}$ be the moles of $KClO_3$
$\frac{n_{KClO_3}}{n_{O_2}}=\frac{2}{3}$
$n_{KClO_3}=\frac{2}{3}\times n_{O_2}=\frac{2}{3}\times0.00178\approx0.00119\space mol$

## Step 5: Find the mass of $KClO_3$
The molar mass of $KClO_3$: $K = 39.10\space g/mol$, $Cl = 35.45\space g/mol$, $O = 16.00\space g/mol$
$M_{KClO_3}=39.10 + 35.45+3\times16.00=122.55\space g/mol$
Mass of $KClO_3$, $m=n_{KClO_3}\times M_{KClO_3}=0.00119\times122.55\approx0.146\space g$

## Step 6: Find the mass percent of $KClO_3$
The mass of the mixture $m_{mixture}=0.7811\space g$
Mass percent of $KClO_3=\frac{mass\space of\space KClO_3}{mass\space of\space mixture}\times100=\frac{0.146}{0.7811}\times100\approx18.7\%$

# Answer:
The mass percent of potassium chlorate in the mixture is approximately $\boldsymbol{18.7\%}$

### 11A. When a sample of zinc reacted with excess hydrochloric acid... (To solve this, we need to follow these steps)

# Explanation:
## Step 1: Write the chemical equation
The reaction between zinc ($Zn$) and hydrochloric acid ($HCl$) is $Zn(s)+2HCl(aq)=ZnCl_2(aq)+H_2(g)$

## Step 2: Find the partial pressure of $H_2$
The atmospheric pressure $P_{atm}=0.963\space atm$, the vapor pressure of water at $30^{\circ}C$ is $P_{H_2O}=32\space torr$. Convert $P_{H_2O}$ to atm: $P_{H_2O}=\frac{32}{760}\space atm\approx0.0421\space atm$
By Dalton's law of partial pressures, $P_{H_2}=P_{atm}-P_{H_2O}$
$P_{H_2}=0.963 - 0.0421 = 0.9209\space atm$

## Step 3: Find the moles of $H_2$ using the ideal gas law
The volume of $H_2$ collected $V = 340\space mL = 0.340\space L$, temperature $T=30^{\circ}C + 273.15=303.15\space K$, $R = 0.0821\space L\cdot atm/(mol\cdot K)$
From $PV=nRT$, $n_{H_2}=\frac{P_{H_2}V}{RT}=\frac{0.9209\times0.340}{0.0821\times303.15}$
$n_{H_2}=\frac{0.3131}{24.89}\approx0.01258\space mol$

## Step 4: Use stoichiometry to find moles of $Zn$
From the balanced equation, the mole ratio of $Zn$ to $H_2$ is $1:1$. So $n_{Zn}=n_{H_2}=0.01258\space mol$

## Step 5: Find the mass of $Zn$
The molar mass of $Zn$ is $65.38\space g/mol$
Mass of $Zn$, $m=n_{Zn}\times M_{Zn}=0.01258\times65.38\approx0.822\space g$

# Answer:
The mass of zinc consumed is approximately $\boldsymbol{0.822\space g}$

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QUESTION IMAGE

Question

Explanation:

13A. Calculate the rms speed of nitrogen dioxide at 85°C.

Step 1: Recall the formula for rms speed

Step 2: Convert temperature to Kelvin

Step 3: Determine the molar mass of \(NO_2\)

Step 4: Substitute values into the formula

Answer:

12B. Does a 1 - mol sample of Ar have the same total KE as a 1 - mol sample of Rn, at a given temperature?