Question

1. what volume would be occupied by 4.8 g of oxygen gas at 0.50 atm and 133°c?

1. what is the molarity of a solution which contains 58.5 g of sodium chloride dissolved in 0.500 l of solution?

1. zn(s) + 2hcl(aq) → h₂(g) + zncl₂(aq)

when 25.0 g of zn reacts, how many l of h₂ gas are formed at stp?

1. complete the unbalanced double replacement equation and indicate if a precipitate forms.

naoh (aq) + mgcl₂(aq) →

1. balance the equation:

nh₃ (g) + o₂ (g) → no₂ (g) + h₂o (g)

Explanation:

Step1: Calculate moles of O₂

Molar mass of $\text{O}_2$ is $32\ \text{g/mol}$.
$n_{\text{O}_2} = \frac{4.8\ \text{g}}{32\ \text{g/mol}} = 0.15\ \text{mol}$

Step2: Convert temperature to Kelvin

$T = 133^\circ\text{C} + 273.15 = 406.15\ \text{K}$

Step3: Use ideal gas law $V=\frac{nRT}{P}$

$R = 0.0821\ \frac{\text{L·atm}}{\text{mol·K}}$, $P=0.50\ \text{atm}$
$V = \frac{0.15\ \text{mol} \times 0.0821\ \frac{\text{L·atm}}{\text{mol·K}} \times 406.15\ \text{K}}{0.50\ \text{atm}} \approx 10.0\ \text{L}$

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Step1: Calculate moles of NaCl

Molar mass of $\text{NaCl}$ is $58.5\ \text{g/mol}$.
$n_{\text{NaCl}} = \frac{58.5\ \text{g}}{58.5\ \text{g/mol}} = 1.00\ \text{mol}$

Step2: Calculate molarity $M=\frac{n}{V}$

$V=0.500\ \text{L}$
$M = \frac{1.00\ \text{mol}}{0.500\ \text{L}} = 2.00\ \text{M}$

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Step1: Calculate moles of Zn

Molar mass of $\text{Zn}$ is $65.38\ \text{g/mol}$.
$n_{\text{Zn}} = \frac{25.0\ \text{g}}{65.38\ \text{g/mol}} \approx 0.382\ \text{mol}$

Step2: Mole ratio of Zn:H₂ is 1:1

$n_{\text{H}_2} = 0.382\ \text{mol}$

Step3: Volume at STP ($22.4\ \text{L/mol}$)

$V_{\text{H}_2} = 0.382\ \text{mol} \times 22.4\ \frac{\text{L}}{\text{mol}} \approx 8.56\ \text{L}$

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Step1: Complete double replacement

Swap cations/anions: $\text{NaOH(aq)} + \text{MgCl}_2\text{(aq)}
ightarrow \text{Mg(OH)}_2 + \text{NaCl}$

Step2: Balance the equation

$2\text{NaOH(aq)} + \text{MgCl}_2\text{(aq)}
ightarrow \text{Mg(OH)}_2\text{(s)} + 2\text{NaCl(aq)}$

Step3: Identify precipitate

$\text{Mg(OH)}_2$ is insoluble in water, so it forms a precipitate.

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Step1: Balance N atoms

$4\text{NH}_3\text{(g)} + \text{O}_2\text{(g)}
ightarrow 4\text{NO}_2\text{(g)} + \text{H}_2\text{O(g)}$

Step2: Balance H atoms

$4\text{NH}_3\text{(g)} + \text{O}_2\text{(g)}
ightarrow 4\text{NO}_2\text{(g)} + 6\text{H}_2\text{O(g)}$

Step3: Balance O atoms

$4\text{NH}_3\text{(g)} + 7\text{O}_2\text{(g)}
ightarrow 4\text{NO}_2\text{(g)} + 6\text{H}_2\text{O(g)}$

Answer:

1. $\approx 10.0\ \text{L}$
2. $2.00\ \text{M}$
3. $\approx 8.56\ \text{L}$
4. $2\text{NaOH(aq)} + \text{MgCl}_2\text{(aq)}

ightarrow \text{Mg(OH)}_2\text{(s)} + 2\text{NaCl(aq)}$; A precipitate of $\text{Mg(OH)}_2$ forms.

1. $4\text{NH}_3\text{(g)} + 7\text{O}_2\text{(g)}

ightarrow 4\text{NO}_2\text{(g)} + 6\text{H}_2\text{O(g)}$