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Slav-nsk [51]
3 years ago
11

Determine the concentration of sulfuric acid that needed 47 mL of 0.39M potassium hydroxide solution to neutralize a 25 mL sampl

e of the sulfuric acid solution?
Chemistry
1 answer:
77julia77 [94]3 years ago
7 0

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
  • M = 0.009165mol/(25mL) × (1,000mL/liter) = 0.3666M

Round to two significant figures: 0.37M

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Answer:

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Explanation:

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Now the ratio of two concentrations is given as

ln (\frac{C}{C_0})=-kt

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So time t is given as

ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs

So the concentration will become 15% of the initial value after 10.31 hrs.

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Answer:

-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

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Mn^2+(aq) + 2e ----> Mn(s)

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ddd [48]

<u>Answer:</u> The amount of heat released is 56 MJ.

<u>Explanation:</u>

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Putting values in above equation, we get:

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Converting this into Mega joules, using the conversion factor:

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