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2 years ago

How many moles are needed

Chemistry

1 answer:

geniusboy [140]2 years ago

7 0

Answer: amount of substance n = cV = 2.0 mol/l x 0.25 l = 0.50 mol

Explanation:

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3 years ago

5. Which of the following materials

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Answer: acetone

Explanation: Acetone has the weakest intermolecular forces, so it evaporated most quickly. Water had the strongest intermolecular forces and evaporated most slowly.

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3 years ago

What is the density of a substance with a mass of 1.8 kg and a volume of 0.2 L?

horrorfan [7]

Answer:

<h3>The answer is 9.0 kg/L</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 1.8 kg

volume = 0.2 L

We have

$density = \frac{1.8}{0.2} \\$

We have the final answer as

Hope this helps you

5 0

3 years ago

How much heat energy is required to raise the temperature of 15g of silver from 25°C to 55°C, assuming the specific heat of silv

Alenkasestr [34]

Answer:

108 j

Explanation:

.24 j/g-C * 15 g * (55-25) =

6 0

2 years ago

When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H

Debora [2.8K]

Answer:

$Y=58.15\%$

Explanation:

Hello,

For the given chemical reaction:

$Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)$

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

$n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3$

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

$n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol Al_2S_3$

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

$m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3$

Finally, we compute the percent yield with the obtained 2.10 g:

$Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%$

Best regards.

7 0

2 years ago