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Phoenix [80]
3 years ago
11

At a pressure of 9.25×10−14 atm and an ordinary temperature of 300.0 K , how many molecules are present in a volume of 1.10 cm3

?
Chemistry
1 answer:
statuscvo [17]3 years ago
7 0

Answer:

2.49 × 10⁶ molecules

Explanation:

Given data

  • Pressure (P): 9.25 × 10⁻¹⁴ atm
  • Temperature (T): 300.0 K
  • Volume (V): 1.10cm^{3} .\frac{1mL}{1cm^{3}} .\frac{1L}{1000mL} =1.10\times 10^{-3} L

We can calculate the moles of gas using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 9.25 × 10⁻¹⁴ atm × 1.10 × 10⁻³ L / (0.0821 atm.L/mol.K) × 300.0 K

n = 4.13 × 10⁻¹⁸ mol

1 mole contains 6.02 × 10²³ molecules (Avogadro's number). The number of molecules in 4.13 × 10⁻¹⁸ moles is:

4.13 × 10⁻¹⁸ mol × (6.02 × 10²³ molecule/1 mol) = 2.49 × 10⁶ molecule

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3 years ago
Determine the concentration of sulfuric acid that needed 47 mL of 0.39M potassium hydroxide solution to neutralize a 25 mL sampl
77julia77 [94]

Answer:

<u></u>

  • <u>0.37M</u>

Explanation:

Since sulfuric acid, H₂SO₄, is a diprotic acid and potassum hydroxide, KOH, contains one OH⁻ in the formula, the number of moles of potassium hydroxide must be twice the number of moles of sulfuric acid.

<u>1. Determine the number of moles of KOH in 47mL of 0.39M potassium hydroxide solution</u>

  • number of moles = molarity × volume in liters
  • number of moles = 0.39M × 47mL × 1liter/1,000 mL = 0.1833mol

<u>2. Determine the number of moles of sulfuric acid needed</u>

  • number of moles of H₂SO₄ = number of moles of KOH/2 = 0.1833/2 = 0.009165mol

<u>3. Determine the concentration that contains 0.009165 mol in 25mL of the acid.</u>

  • Molarity = number of moles / volume in liters
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3 years ago
A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
larisa86 [58]

Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

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