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Y_Kistochka [10]
3 years ago
6

What volume of 0.20 M NaCl contains 10.0 g

Chemistry
1 answer:
nalin [4]3 years ago
8 0
0.862 liters of a 0.20 M NaCl solution (aq) contains 10.0g of NaCl.
Answer : 0.862 liters
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Condensation is a process for which a. ΔG is negative at high temperature but positive at low temperature. b. ΔH and ΔS are nega
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Answer:

b. ΔH and ΔS are negative at all temperatures .

Explanation:

During the process of condensation ,

The gaseous state convert to liquid state ,

Hence , the entropy of the system reduces , i.e. , the randomness decreases .

And the value for entropy is negative ,

hence ,

Δ S = negative ,

Δ H = negative ,

Since ,

The heat is releasing from system .

hence , the most appropriate option will be ΔH and ΔS are negative at all temperatures .

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What color does copper produce when it reacts to chlorine
kiruha [24]

The color it produces is Bright green.

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4.5x10'25 atoms of nickel equal how many moles
olga2289 [7]
1 mole of any particles = 6.02* 10²³ particles

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8 0
3 years ago
How do I do this? What are the answers to the 5 questions shown?
frozen [14]

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

6 0
3 years ago
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