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dalvyx [7]
3 years ago
13

A potassium bromide solution is 8.30% potassium bromide by mass and its density is 1.03 g/ml. what mass of potassium bromide is

contained in 41.2 ml of the solution?
Chemistry
1 answer:
pshichka [43]3 years ago
7 0

<u>Answer:</u> The mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

<u>Explanation:</u>

We are given that KBr is present in 8.3% KBr solution, which means that 8.3 grams of potassium bromide is present in 100 gram of the solution.

To calculate the volume of KBr, we use the formula:

Density=\frac{Mass}{Volume}

Mass of the solution = 100 grams

Density of KBr solution = 1.03g/mL

Volume of the solution = ? mL

Putting values in above equation, we get:

1.03g/mL=\frac{100g}{Volume}\\\\Volume=97.08mL

Now, to calculate the mass of KBr in 41.2mL of the solution, we use unitary method.

In 97.08 mL of solution, mass of KBr present is 8.3 grams.

So, 41.2 mL of solution will contain = \frac{8.3g}{97.08mL}\times 41.2mL=3.522g of KBr.

Hence, the mass of potassium bromide present in 41.2 mL of solution will be 3.522 grams.

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Answer:

Fe₂(SO₄)₃ + 6KOH —> 3K₂SO₄ + 2Fe(OH)₃

The coefficients are: 1, 6, 3, 2

Explanation:

__Fe₂(SO₄)₃ + __KOH —> __K₂SO₄ + __Fe(OH)₃

To determine the correct coefficients, we shall balance the equation. This can be obtained as follow:

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There are 2 atoms of Fe on the left side and 1 atom on the right side. It can be balance by writing 2 before Fe(OH)₃ as shown below:

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There are 6 atoms of OH on the right side and 1 atom on the left side. It can be balance by writing 6 before KOH as shown below:

Fe₂(SO₄)₃ + 6KOH —> K₂SO₄ + 2Fe(OH)₃

There are 6 atoms of K on the left side and 2 atoms on the right side. It can be balance by writing 3 before K₂SO₄ as shown below:

Fe₂(SO₄)₃ + 6KOH —> 3K₂SO₄ + 2Fe(OH)₃

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