Assume an object A with a mass of 66.789 g and a volume of 10.1 mL. what is its density?

Help me out ! What’s the answer

Anni [7]

Answer:

true

Explanation:

5 0

2 years ago

the two stages that take place during distillation are. a.evaporation is folloeed by condensation. b.condensation followed by ev

ValentinkaMS [17]

Answer:

evapiration followed by condensation

3 0

2 years ago

The neutrons within the nucleus of an atom have what charge?

vivado [14]

Answer:

the answer is D.

Explanation:

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4 0

2 years ago

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Korolek [52]

Answer:

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Explanation:

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6 0

2 years ago

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO 3 ( s ) + 2

e-lub [12.9K]

Answer: The mass of calcium chloride formed is 15.21 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For calcium carbonate:

Given mass of calcium carbonate = 32.0 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium carbonate}=\frac{32.0g}{100g/mol}=0.32mol$

For hydrochloric acid:

Given mass of hydrochloric acid = 10.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrochloric acid}=\frac{10.0g}{36.5g/mol}=0.274mol$

The given chemical equation follows:

$CaCO_3(s)+2HCl(aq.)\rightarrow CaCl_2(aq.)+H_2O(l)+CO_2(g)$

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium carbonate

So, 0.274 moles of hydrochloric acid will react with = $\frac{1}{2}\times 0.274=0.137mol$ of calcium carbonate

As, given amount of calcium carbonate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.274 moles of hydrochloric acid will produce = $\frac{1}{2}\times 0.274=0.137moles$ of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 111 g/mol

Moles of calcium chloride = 0.137 moles

Putting values in equation 1, we get:

$0.137mol=\frac{\text{Mass of calcium chloride}}{111g/mol}\\\\\text{Mass of calcium chloride}=(0.137mol\times 111g/mol)=15.21g$

Hence, the mass of calcium chloride formed is 15.21 grams.

4 0

2 years ago