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vredina [299]
3 years ago
15

Calculate the number of moles in 15.5g of CaSO4.5H2O

Chemistry
1 answer:
Natali [406]3 years ago
4 0

Answer:

No. of moles = 0.0685

Explanation:

Given mass, m = 15.5g

We need to find the number of moles in 15.5 g of CaSO₄.5H2O

First, we find the mass of CaSO₄.5H2O.

M = (1×40)+(1×32)+(4×16)+(5×18)

M = 226 g/mol

We know that,

Number of moles = given mass/molar mass

n=\dfrac{15.5\ g}{226\ g/mol}\\\\n=0.0685\ \text{mol}

Hence, there are 0.0685 moles in 15.5 g of CaSO₄.5H2O.

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Color change
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Put the list in chronological order (1–5).
Leokris [45]

Explanation:

Filtration is a separation technique in which solid particles suspended in liquid medium are separated by allowing the mixture through the pores of the filter paper. By this solid particles get collect on filter paper and liquid drains out from the pores of the filter paper.

The chronological order for given steps will be:

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3 years ago
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100 POINTS PLEASE HELP!! Honors Stoichiometry Activity Worksheet Instructions: In this laboratory activity, you will taste test
Shtirlitz [24]

Answer:

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=

236.59g/mol

946.36g

=4mol

Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=

225g/mol

196.86g

=0.8749mol

Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=

257.83g/mol

193.37g

=0.7499mol

Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=

719.42g/mol

2050.25g

=2.8498mol

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

\frac{2}{1}\times 0.7499 mol = 1.4998 mol

1

2

×0.7499mol=1.4998mol of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

\frac{1}{1}\times 0.7499 mol = 0.7499 mol

1

1

×0.7499mol=0.7499mol of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

\frac{4}{1}\times 0.7499 mol=2.996 mol

1

4

×0.7499mol=2.996mol of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100

theoretical yield

Experimental yield

×100

\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%

2157.9722g

2050.25g

×100=95.00%

6 0
3 years ago
Read 2 more answers
What is the molarity of a solution that contains 0.400 mol HCI in 9.79 L solutions ​
USPshnik [31]

Molarity's formula is known as: Molarity(M)=moles of solute/liters solution.

In this case we are already given moles and liters so you just have to plug the numbers into the equation.

0.400 mol HCL/9.79L solution=0.040858M

If you were to use scientific notation, the answer will be: 4.1*10^-2, but otherwise, you can just use the decimals above and round appropriately as you see fit.

6 0
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Answer:

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Explanation:

Scientific experiments have concluded that it takes approximately 23 days for the moon to rotate and also it takes the same duration for the moon to revolve around the Earth. Due to this consistency, the moon appears to be still.

<em>Such synchronization results in the same face of the moon to be directed towards the Earth. Hence, the same craters of the moon will be observed by the scientist every day.</em>

<em></em>

Other options, like option D, is not correct because there will be craters on the other side of the moon too. But as we see the same side of the moon, hence we cannot see the craters present on the other side of the moon.

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3 years ago
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