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2 years ago

What mass of 2-bromopropane could be prepared from 25.5 g of propene? Assume a 100% yield of product.

Chemistry

1 answer:

Mamont248 [21]2 years ago

5 0

Answer:

78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene

Explanation:

$C_3H_6+HBr\rightarrow C_3H_7Br$

Moles of propene = $\frac{25.5 g}{39 g/mol}=0.6538 mol$

According to reaction, 1 mole of propene gives 1 mole of propane.

Then 0.6538 moles of bromo-propane will give:

$0.6538 mol\times 120 g/mol=78.46 g$

78.46 grams of 2-bromopropane could be prepared from 25.5 g of propene.

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How many iodide ions are present in 65.5ml of .210 m AlI3 solution

blondinia [14]

Answer:

$2.48\times 10^{22} ions$ are present in solution.

Explanation:

Molarity of the solution = 0.210 M

Volume of the solution = 65.5 ml = 0.0655 L

Moles of aluminum iodide= n

$Molarity=\frac{\text{Moles of compound}}{\text{Volume of the solution(L)}}$

$0.210M=\frac{n}{0.0655 L}$

n = 0.013755 moles of aluminum iodide

1 mole of aluminum iodide contains 3 moles of iodide ions:

Then 0.013755 moles of aluminum iodide will contain:

$3\times 0.013755 moles=0.041265 mol$ of iodide ions

Number of iodide ions in 0.041265 moles:

$0.041265 mol\times 6.022\times 10^{23}=2.48\times 10^{22} ions$

$2.48\times 10^{22} ions$ are present in solution.

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2 years ago

Among these processes, which is the slowest chemical reaction?

user100 [1]

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djverab [1.8K]

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