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jek_recluse [69]
3 years ago
6

Given that NASA needs 151,200 g of Oxygen (O2) per person for a mission, how many grams of water (H2O) does NASA need to ship to

meet the Oxygen demands? Use Stoichiometry to determine your answer. SHOW YOUR WORK FOR THIS 3-STEP PROBLEM. Make sure to balance the equation first. H2O (l) → H2 (g) + O2 (g)
Chemistry
1 answer:
IgorC [24]3 years ago
6 0

this is so ez,

ok so here what u'll do after u balance the equation:

1. convert grams of oxygen to moles of oxygen.

2. convert moles of oxygen to moles of water

3. convert moles of water to grams of water.

4. bOOm... that's your answer

as follows:

2H2O(l)===>2H2(g)+O2(g)

1. 151200(g)/16(g/mol)=9540 moles of O2

2. 9450 moles of O2 × 2 moles of H2O =18900 moles of H2O

3. 18900 moles of H2O × 10g/mol = 189000g of H2O is required.

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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}

p_v=14mmHg

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

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\text{Percent humidity}=79.62\%

Therefore, the percent humidity is 79.62 %

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