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2 years ago

9. Suppose that 25.0 mL of a gas at 725 mm Hg and 20°C is converted to standard

Chemistry

1 answer:

storchak [24]2 years ago

8 0

Answer:

V₂ = 22.23 mL

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 25 mL

Initial pressure = 725 mmHg (725/760 =0.954 atm)

Initial temperature = 20 °C (20 +273 = 293 K)

Final pressure = standard = 1 atm

Final temperature = standard = 273.15 K

Final volume = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ = 0.954 atm × 25 mL × 273.15 K / 293 K × 1 atm

V₂ = 6514.63 mL . atm . K / 293 K . atm

V₂ = 22.23 mL

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vladimir2022 [97]

Answer:

The solution's new volume is 1.68 L

Explanation:

Dilution is the procedure to prepare a less concentrated solution from a more concentrated one, and simply consists of adding more solvent. So, in a dilution the amount of solute does not vary, but the volume of the solvent varies.

In summary, a dilution is a lower concentration solution than the original.

The way to do the calculations in a dilution is through the expression:

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Ci= 7 M
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7 M*0.60 L= 2.5 M* Vf

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3 years ago

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How many grams of hydrogen chloride can be produced from 1.00g of hydrogen and 55.0g of chlorine? what is the limiting reactant?

zloy xaker [14]

Answer:

$m_{HCl}=36.1gHCl$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required grams of HCl by firstly identifying the limiting reactant via the moles of each reactant as they are in a 1:1 mole ratio:

$n_{H_2}=1.00gH_2*\frac{1molH_2}{2.02gH_2}=0.500molH_2\\\\ n_{Cl_2}=55.0gCl_2*\frac{1molCl_2}{70.9gCl_2}=0.776molCl_2$

Thus, we infer the hydrogen is the limiting reactant and therefore we use its 1:2 mole ratio with HCl whose molar mass is 36.46 g/mol:

$m_{HCl}=0.500molH_2*\frac{2molHCl}{1molH_2}*\frac{36.46gHCl}{1molHCl}\\\\m_{HCl}=36.1gHCl$

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2 years ago