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3 years ago

PLEASE ANSWER ASAP

Chemistry

2 answers:

RSB [31]3 years ago

5 0

Question:

What would be the best way to increase the rate

of the reaction?

Answer:

D. She could make the original solutions at higher concentrations.

Llana [10]3 years ago

3 0

Answer:

D. She could make the original solutions at higher concentrations.

Explanation:

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How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 30.6 moles of water vapor?

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1 mol H₂ - 1 mol H₂O
x mol H₂ - 30.6 mol H₂O

x=30.6*1/1=30.6 mol (<span>theoretically)</span>

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4 years ago

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Hydrogen bonds are a strong attractive force, since they are less stronger than ionic bonds.

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Freeze-drying is a process used to preserve food. If strawberries are to be freeze-dried, then they would be frozen to -80.00 °C

katrin2010 [14]

Answer:

1. 389 kJ; 2. 7.5 µg; 3. 6.25 days

Explanation:

1. Energy required

The water is converted directly from a solid to a gas (sublimation).

They don't give us the enthalpy of sublimation, but

$\Delta_{\text{sub}}H = \Delta_{\text{fus}}H + \Delta_{\text{vap}}H = 6.01 + 40.68 = 46.69 \text{ kJ}\cdot\text{mol}^{-1}$

The equation for the process is then

Mᵣ: 18.02

46.69 kJ + H₂O(s) ⟶ H₂O(g)

m/g: 150

(a) Moles of water

$\text{Moles} = \text{150 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{8.324 mol}$

(b) Heat removed

46.69 kJ will remove 1 mol of ice.

$\text{Heat removed} = \text{8.234 mol} \times \dfrac{\text{46.69 kJ}}{\text{1 mol}} = \textbf{389 kJ}\\\text{It takes $\large \boxed{\textbf{389 kJ}}$ to remove 150 g of ice}$

2. Mass of water vapour in the freezer

For this calculation, we can use the Ideal Gas Law — pV = nRT

(a) Moles of water

Data:

$p = 1.00 \times 10^{-3}\text{ torr } \times \dfrac{\text{1 atm}}{\text{760 torr}} = 1.316 \times 10^{-6}\text{ atm}$

V = 5 L

T = (-80 + 273.15) K = 193.15 K

Calculation:

$\begin{array}{rcl}pV & = & nRT\\1.316 \times 10^{-6}\text{ atm} \times \text{5 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{193.15 K }\\6.6 \times 10^{-6} & = & 15.85n\text{ mol}^{-1} \\n & = & \dfrac{6.6 \times 10^{-6}}{15.85\text{ mol}^{-1}}\\\\& = & 4.2 \times 10^{-7} \text{ mol}\\\end{array}$

(b) Mass of water

$\text{Mass} = 4.2 \times 10^{-7} \text{ mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = 7.5 \times 10^{-6}\text{ g} = 7.5 \, \mu \text{g}\\\\\text{At any given time, there are $\large \boxed{\textbf{7.5 $\mu$g}}$ of water vapour in the freezer.}$

3. Time for removal

You must remove 150 mL of water.

It takes 1 h to remove 1 mL of water.

$\text{Time} = \text{150 mL} \times \dfrac{\text{1 h}}{\text{1 mL}} = \text{150 h} = \text{6.25 days}$

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4 years ago