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3 years ago

What is the ph of a 0.006 m koh solution? 1. 11.78 2. 7.00 3. 2.22 4. 8.88 5. 5.12?

1 answer:

4 0

The concentration of OH⁻ is first converted to pOH bu using followinf formula,

pOH = -log [OH⁻]
Putting value,
pOH = -log (0.006)

pOH = 2.221
As we know,
pH + pOH = 14
Solving for pH,
pH = 14 - pOH
Putting value of pOH,
pH = 14 - 2.221

pH = 11.779
Result:
Option-1 is the correct answer.

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The empirical formula of the organic compound is = $C_2H_6S_1$

Explanation:

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Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of $SO_2$ gas at STP occupying 1.0 L = n''

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$n''=\frac{1.0 L}{22.4 L}=0.04464 mol$

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

Moles of carbon , hydrogen and sulfur constituent of that organic compound .

Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol

Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol

Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol

For empirical; formula divide the least number of moles from all the moles of elements.

carbon = $\frac{0.08920 mol}{0.04464 mol}=2$

Hydrogen = $\frac{0.2678 mol}{0.04464 mol}=6$

Sulfur = $\frac{0.04464 mol}{0.04464 mol}=1$

The empirical formula of the organic compound is = $C_2H_6S_1$

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2 years ago

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Answer:

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Explanation:

<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>

Step 1: Given data

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Initial volume (V₁): 230. mL

Final concentration (C₂): ?

Final volume (V₂): 660. mL

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