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iris [78.8K]
3 years ago
7

How many moles are equal to a 0.25g sample of chromium?

Chemistry
1 answer:
jonny [76]3 years ago
7 0

Answer:

0.0048 mol Cr

Explanation:

Step 1: Define

Molar Mass of Chromium (Cr) - 52.00 g/mol

Step 2: Use Dimensional Analysis

0.25 \hspace{3} g \hspace{3} Cr(\frac{1 \hspace{3} mol \hspace{3} Cr}{52.00 \hspace{3} g \hspace{3} Cr} ) = 0.004808 mol Cr

Step 3: Simplify

We are give 2 sig figs.

0.004808 mol Cr ≈ 0.0048 mol Cr

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Answer:

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Explanation:

From the collision theory, we know that it is super improbable for 3 different molecules (2 NO and 1 O2) to all hit each other at the perfect speed in the perfect position to make the products. From this, we can pretty confidently say that this is most likely a multi-stepped reaction.

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Match each of the three descriptions of a volume to the appropriate metric unit of volume
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2 years ago
9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r
Mumz [18]

Answer

b. 95%

Explanation

Given:

Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\  \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\  \\ Percent\text{ }yield=0.9484\times100\% \\  \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}

Therefore, the percent yield for the reaction is 95%.

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