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2 years ago

Calculate the new volume of 1.23 mL of a gas at 32 C is subjected to drop in temperature of 20 degrees Celsius

Chemistry

1 answer:

kotykmax [81]2 years ago

8 0

Answer:

1,15mL = V₂

Explanation:

Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:

V₁T₂ = V₂T₁

Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.

The V₁ is 1.23mL

T₁ = 32°C + 273.15 = 305.15K

T₂ = T₁ - 20°C = 285.15K

Replacing:

1.23mL*285.15K = V₂*305.15K

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How many molecules of O2 are in 153.9 g O2?

MrRissso [65]

Answer:

2.895*10^24

Explanation:

mass of Oxygen give = 153.9g

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3 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

The nucleus of most atoms is composed of which of the following sub-atomic particles?

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Answer:

C. Tightly packed protons and neutrons

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It consists of two sub-atomic particle namely proton and neutron.

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8 0

2 years ago

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