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2 years ago

12

A balloon contains 7. 2 L of helium. The pressure is reduced to 2. 00 atm and the balloon expands to occupy a volume of 25. 1 L.

What was the initial pressure exerted on the balloon?

1 answer:

PSYCHO15rus [73]2 years ago

3 0

Answer:

your answer should be 0.57 atm

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Air at 293 K and 750 mm Hg pressure has a relative humidity of 80%. What is its percent humidity? The vapour pressure of water a

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Answer : The correct option is, (c) 79.62

Explanation :

The formula used for percent humidity is:

$\text{Percent humidity}=\text{Relative humidity}\times \frac{p-p^o_v}{p-p_v}\times 100$ ..........(1)

The formula used for relative humidity is:

$\text{Relative humidity}=\frac{p_v}{p^o_v}$ ...........(2)

where,

$p_v$ = partial pressure of water vapor

$p^o_v$ = vapor pressure of water

p = total pressure

First we have to calculate the partial pressure of water vapor by using equation 2.

Given:

$p=750mmHg$

$p^o_v=17.5mmHg$

Relative humidity = 80 % = 0.80

Now put all the given values in equation 2, we get:

$0.80=\frac{p_v}{17.5mmHg}$

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Now we have to calculate the percent humidity by using equation 1.

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4 years ago

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Explanation:

We must convert formula units of Zn(ClO₃)₂ to moles and then to grams of Zn(ClO₃)₂.

Step 1. Convert formula units to moles

$\text{Moles of Zn(ClO$_{3}$)$_{2}$}\\\\= 9.80 \times10^{23}\text{ formula units Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}{6.022 \times\ 10^{23} \text{ formula units Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$}$

Step 2. Convert moles to grams

$\text{Mass of Zn(ClO$_{3}$)$_{2}$}\\\\= \text{1.627 mol Zn(ClO$_{3}$)$_{2}$} \times \dfrac{\text{232.29 g Zn(ClO$_{3}$)$_{2}$}}{\text{1 mol Zn(ClO$_{3}$)$_{2}$}}\\\\= \text{378 g Zn(ClO$_{3}$)$_{2}$}\\\\\text{The mass of Zn(ClO$_{3}$)$_{2}$ is } \boxed{\textbf{378 g}}$

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Answer:

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I hope I helped! ^-^

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