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3 years ago

(QUICK) Which statement is true about percent yield?

Chemistry

1 answer:

jek_recluse [69]3 years ago

8 0

Answer: the answer is A.

Explanation:

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PLEASE ANSWER + BRAINLIEST !!!

aleksklad [387]

(bottom 36)(top 81) Kr

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4 years ago

How many inches are in 217.1 centimeters?

dusya [7]

Answer:

85.47

Explanation:

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4 years ago

A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4. (a) Write the balanced chemical equ

NISA [10]

Answer:

a) 2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂

b) Ni(OH)₂

c) KOH

d) 0.927 g

e) K⁺=0.067 M, SO₄²⁻=0.1 M, Ni²⁺=0.067 M

Explanation:

a) The equation is:

2KOH + NiSO₄ → K₂SO₄ + Ni(OH)₂ (1)

b) The precipitate formed is Ni(OH)₂

c) The limiting reactant is:

$n_{KOH} = V*M = 100.0 \cdot 10^{-3} L*0.200 mol/L = 0.020 moles$

$n_{NiSO_{4}} = V*M = 200.0 \cdot 10^{-3} L*0.150 mol/L = 0.030 moles$

From equation (1) we have that 2 moles of KOH react with 1 mol of NiSO₄, so the number of moles of KOH is:

$n = \frac{2}{1}*0.030 moles = 0.060 moles$

Hence, the limiting reactant is KOH.

d) The mass of the precipitate formed is:

$n_{Ni(OH)_{2}} = \frac{1}{2}*n_{KOH} = \frac{1}{2}*0.020 moles = 0.010 moles$

$m = n*M = 0.010 moles*92.72 g/mol = 0.927 g$

e) The concentration of the SO₄²⁻, K⁺, and Ni²⁺ ions are:

$C_{K^{+}} = \frac{2*\frac{1}{2}*n_{KOH}}{V} = \frac{0.020 moles}{0.300 L} = 0.067 M$

$C_{SO_{4}^{2-}} = \frac{\frac{1}{2}*n_{KOH + (0.03 - 0.01)}}{V} = \frac{0.030 moles}{0.300 L} = 0.1 M$

$C_{Ni^{2+}} = \frac{0.020 moles}{0.300 L} = 0.067 M$

I hope it helps you!

5 0

4 years ago

What is the significance of a gene?

Zanzabum

Answer : DNA (if im wrong please correct me)

3 0

3 years ago

You have a sample that has a h3o+ concentration of 1.25 x 10-5 m. what would be the [oh-]?

deff fn [24]

Answer:

[OH⁻] = 8x10⁻¹⁰

Explanation:

Kw = 1x10⁻¹⁴

Kw = [H₃O⁺] . [OH⁻]

1x10⁻¹⁴ = 1.25x10⁻⁵ . [OH⁻]

1x10⁻¹⁴ / 1.25x10⁻⁵ = [OH⁻]

[OH⁻] = 8x10⁻¹⁰

7 0

4 years ago