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bixtya [17]
3 years ago
7

Helium gas in a cylinder is under 1.12atm pressure at 25.0C. What will be the pressure if the temperature increases to 37.0C?

Chemistry
1 answer:
Irina18 [472]3 years ago
7 0

Answer:

p_2=1.17atm

Explanation:

Hello!

In this case, considering that the Gay-Lussac's law allows us to relate the temperature-pressure problems as directly proportional relationships:

\frac{p_2}{T_2} =\frac{p_1}{T_1} \\\\

Thus, for the initial pressure and temperature in kelvins the final temperature in kelvins, we compute the final pressure as:

p_2=\frac{p_1T_2}{T_1} \\\\p_2=\frac{1.12atm*310.15K}{298.15K}\\\\p_2=1.17atm

Best regards!

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Suppose 2.4 g of Mg reacts with 10.0 g of O2, to create magnesium oxide. How much magnesium oxide is produced?
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Answer:

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Explanation:

Step 1. Write the balanced chemical reaction. In this case, magnesium reacts with oxygen to produce magnesium oxide:

2~Mg(s) + O_2 (g)\rightarrow 2~MgO (s)

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Step 2. Calculate the number of moles of magnesium:

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Step 3. Calculate the number of moles of oxygen:

n_{O_2} = \frac{10.0~g}{32.00~g/mol} = 0.3125~mol

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Step 4. Identify the limiting reactant comparing the equivalents. Equivalent of Mg:

eq_{Mg} = \frac{0.0987~mol}{1} = 0.0987~mol

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eq_{O_2} = \frac{0.3125~mol}{2} = 0.15625~mol

Therefore, Mg is the limiting reactant.

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Step 5. According to the stoichiometry of this reaction:

n_{Mg} = n_{MgO} = 0.0987~mol

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Step 6. Convert the number of moles of MgO into mass:

m_{MgO} = 0.0987~mol\cdot 40.304~g/mol = 3.98~g

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