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lbvjy [14]
3 years ago
8

A 44% wt/wt solution of H2SO4 has a density of 1.343g/ml. What mass of H2SO4 is 60ml of this solution?

Chemistry
1 answer:
horsena [70]3 years ago
3 0

<u>Answer:</u> The mass of sulfuric acid present in 60 mL of solution is 34.1 grams

<u>Explanation:</u>

We are given:

44 % (m/m) solution of sulfuric acid. This means that 44 grams of sulfuric acid is present in 100 grams of solution.

To calculate volume of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.343 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

1.343g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.343g/mL}=77.46mL

To calculate the mass of sulfuric acid present in 60 mL of solution, we use unitary method:

In 77.46 mL of solution, mass of sulfuric acid present is 44 g

So, in 60 mL of solution, mass of sulfuric acid present will be = \frac{44g}{77.46mL}\times 60mL=34.1g

Hence, the mass of sulfuric acid present in 60 mL of solution is 34.1 grams

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Use the ICE table approach as solution:

           PbSO₄   --> Pb²⁺ + SO₄²⁻
I             -                 0          0
C           -                +s         +s
E           -                  s          s

Ksp = [Pb²⁺][SO₄²⁻]
1.82×10⁻⁸ = s²
Solving for s,
s = <em>1.35×10⁻⁴ M</em>
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How many grams of sulfur are required in the preparation of 35.7 moles of sulfur dioxide Sg + O2 SO2​
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  • S+O_2-->SO_2

Its balanced already

So

  • 1 mol SO_2 require 1 mole sulphur
  • 35.7moles require 35.7mol sulphur

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  • 1142.4g
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The _________ indicates that 70% of the Sun's radiation is contained on Earth and 30% is reflected back into space.
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g 304 mL of a 0.36 M potassium hydroxide solution is added to 341 mL of a 0.51 M lithium hydroxide solution. Calculate the pOH o
Liono4ka [1.6K]

Answer:

<u><em></em></u>

  • <u><em>pOH = 0.36</em></u>

Explanation:

Both <em>potassium hydroxide</em> and <em>lithium hydroxide </em>solutions are strong bases, so you assume 100% dissociation.

<u>1. Potassium hydroxide solution, KOH</u>

  • Volume, V = 304 mL = 0.304 liter
  • Molarity, M = 0.36 M
  • number of moles, n = M × V = 0.36M × 0.304 liter = 0.10944 mol
  • 1 mole of KOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.10944

<u>2. LIthium hydroxide, LiOH</u>

  • Volume, V = 341 mL = 0.341 liter
  • Molarity, M = 0.51 M
  • number of moles, n = M × V = 0.341 liter × 0.51 M = 0.17391 mol
  • 1mole of LiOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.17391

<u />

<u>3. Resulting solution</u>

  • Number of moles of OH⁻ ions = 0.10944 mol + 0.17391 mol = 0.28335 mol

  • Volume of solution = 0.304 liter + 0.341 liter = 0.645 liter

  • Molar concentration = 0.28335 mol / 0.645 liter = 0.4393 M

<u />

<u>4. </u><em><u>pOH</u></em>

      pOH=-\log [OH^-]=-\log (0.439)=0.36   ← answer

5 0
3 years ago
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