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jok3333 [9.3K]
3 years ago
11

Check all reasons why a book was used.

Chemistry
1 answer:
marin [14]3 years ago
4 0

We can confirm that for the experiments regarding Boyle's law, the book was used to increase pressure on the gas and add additional weight.

<h3>Why is the book used in these experiments?</h3>

The book is used for the reasons described in options 1 and 2 of the question. The book supports additional weights on the container, meaning that through these added weights we can increase the pressure being applied to the gas and see how said pressure affects the volume of a gas.

Therefore, we can confirm that the book was used to increase pressure on the gas and add additional weight.

To learn more about Boyle's law visit:

brainly.com/question/1437490?referrer=searchResults

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How many moles of H2 can be made from complete reaction of 3.0 moles of Al?
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Answer: 4.5 moles of H_{2} can be made from complete reaction of 3.0 moles of Al.

Explanation:

The given reaction equation is as follows.

2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}

This shows that 2 moles of Al reacts with 6 moles of HCl. So, the amount of HCl required to react with 1 mole Al is three times the amount of HCl.

Therefore, 3 moles of Al will react with 9 moles of HCl to give 3 moles of AlCl_{3} and \frac{9}{2} moles of H_{2}.

The reaction equation now will be as follows.

3Al + 9HCl \rightarrow 3AlCl_{3} + \frac{9}{2}H_{2}

The moles \frac{9}{2} can also be written as 4.5 moles.

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3 years ago
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A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i
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81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>

<em />

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

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<em>Moles CaCl₂ = Moles CaCO₃:</em>

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

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