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Oduvanchick [21]
3 years ago
7

exactly 5.00 L of air at -50.0 °C is warmed to 100.0 °C. What is the new volume if the pressure remains constant?

Chemistry
1 answer:
ollegr [7]3 years ago
3 0

Answer:

V₂ = 8.36 L

Explanation:

Given data:

Initial volume of gas = 5.00 L

Initial temperature = -50.0°C (-50 + 273 = 223 k)

Final temperature = 100°C (100+273 = 373 k)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 5.00 L × 373 K / 223 K

V₂ = 1865 L.K / 223 K

V₂ = 8.36 L

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Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions. If 0.212 g of KHP are dissolve
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0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.

Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.

The balanced neutralization equation is:

NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)

  • Step 1: Calculate the reacting moles of KHP.

0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.

0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol

  • Step 2: Determine the reacting moles of NaOH.

The molar ratio of NaOH to KHP is 1:1.

1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH

  • Step 3: Calculate the molarity of NaOH.

1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.

[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M

0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.

Learn more about titration here: brainly.com/question/4225093

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The number of moles of solute is equal to product of the molar concentration (molarity) and the volume (in liters) of solution.

Since the volumes and the molar concentrations of the<em> NaOH </em>and <em>HCl </em>solutions mixed are equal, each one of them contributes the same number of moles of solute.

Since every mol of NaOH produces one mol of OH⁻ ions and every mol of HCl produces one mol of H⁺ ion, the number of moles of OH ⁻ and H⁺ in solution are equal.

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