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3 years ago

14

What is the mass of solute dissolved in 10.0 g of 5.00% Sugar solution?

1 answer:

cupoosta [38]3 years ago

3 0

Answer:

0.5 g or D

Explanation:

what is the mass of solute dissolved in 10.0 g of 5.00% Sugar solution?

A. 0.900 g

B. 0.180 g

C. 10.0 g

D. 0.500 g

E. 9.50 g

5% of the solution is sugar

o 10 g of the solution will have 9.5 g H2O and 0.5 g of sugar

0.5 /(9.5 +0.5) = 0.5/10 = 0.05 = 5 PER HUNDRED = 5%

so 0.5 g or D

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General Chemistry fourth edition by McQuarrie, Rock, and Gallogly. University Science Books presented by Macmillan Learning.

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Answer:

3.07 Cal/g

Explanation:

Step 1: Calculate the heat absorbed by the calorimeter

We will use the following expression.

Q = C × ΔT

where,

Q: heat absorbed

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According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.

Step 2: Convert 86.1 kJ to Cal

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86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal

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Read 2 more answers

HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element i

OleMash [197]

<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

<u>Explanation:</u>

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

$HNO_3+S\rightarrow H_2SO_4+NO$

<u>On the reactant side:</u>

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

<u>On the product side:</u>

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

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2 years ago