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Tom [10]
1 year ago
7

Why is the vapor pressure of a warm pale higher than the vapor pressure of a cold lake

Chemistry
1 answer:
levacccp [35]1 year ago
3 0

The vapor pressure of a warm pale higher than the vapor pressure of a cold lake because Warm water evaporates more quickly.

Hence, Option B is correct answer.

<h3>What is Evaporation ? </h3>

The process in which a liquid or solid is converted into vapour is called Evaporation. When water gains the heat energy then it changes in to gas. When the temperature increases it speeds up the rate of evaporation.

The temperature of warm water increases, the kinetic energy of the warm water molecule also increases.

Thus from the above conclusion we can say that The vapour pressure of a warm lake is higher than the vapor pressure of a cold lake because Warm water evaporates more quickly.

Hence, Option B is correct answer.

Learn more about the Evaporation here: brainly.com/question/9339710

#SPJ1

Disclaimer: The question was given incomplete on the portal. Here is the complete questions.

Question: Why is the vapor pressure of a warm lake higher than the vapor pressure of a cold lake?

A. Warm water has a greater heat of vaporization.

B. Warm water evaporates more quickly.

C. Cool water evaporates more quickly.

D. Cool water has a greater heat of vaporization.

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Answer:

\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}

Explanation:

1. Miles travelled in an average month  

\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}

2. Using a gasoline powered vehicle  

(a) Moles of heptane used  

n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}

(b) Equation for combustion  

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O  

(c) Moles of CO₂ formed  

n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}

(d) Volume of CO₂ formed  

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.  

V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}

3. Using an electric vehicle  

(a) Theoretical energy used  

\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}

(b) Actual energy used  

The power station is only 85 % efficient.  

\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\

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4. Comparison  

\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}

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3 years ago
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Explanation:

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Let me know if this helps, if not I'll continue researching.

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3 years ago
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