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AlexFokin [52]
2 years ago
14

Consider the hypothetical reaction 3A + 4B → C + 2D Over an interval of 2.50 s the average rate of change of the concentration o

f A was measured to be -0.0560 M/s. What is the final concentration of B at the end of this same interval if its concentration was initially 1.700 M?
Chemistry
1 answer:
faust18 [17]2 years ago
3 0

Answer:

Final [B] = 1.665 M

Explanation:

3A + 4B → C + 2D

Average rection rate = 3[A]/Δt = 4[B]/Δt = [C]/Δt = 2[D]/Δt

0.05600 M/s = 4 [B]/ 2.50 s

[B] = 0.035 M (concentration of B consumed)

Final [B] = initial [B] - consumed [B]

Final [B] = 1.700 M - 0.035 M

Final [B] = 1.665 M

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Answer:

Explanation:

Volume is defined as the space occupied by an object or substance irrespective of its state of matter.The conversion used from millimeter to liter is:

1 milliiliter = 0.001 L

Therefore, we can convert the volume of sample from 2.5 ml in liters as follows.

2.5 ml in liters = 2.5ml x 0.001 L/1ml

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2 years ago
Which of the following describes the stage that occurs after condensation in the water cycle?
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Answer:

Precipitation

Explanation:

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2 years ago
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nevsk [136]

Answer:

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2 years ago
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HNO3 + S --&gt; 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

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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
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Elements in the same group have D. Same number of valence electrons.
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