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Sergeeva-Olga [200]
3 years ago
14

A first-order reaction (A → B) has a half-life of 25 minutes. If the initial concentration of A is 0.900 M, what is the concentr

ation of B after 50 minutes?
Chemistry
1 answer:
Oksanka [162]3 years ago
7 0

Answer:

0.6749 M is the concentration of B after 50 minutes.

Explanation:

A → B

Half life of the reaction = t_{1/2}=25 minutes

Rate constant of the reaction = k

For first order reaction, half life and half life are related by:

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{25 min}=0.02772 min^{-1}

Initial concentration of A = [A]_o=0.900 M

Final concentration of A after 50 minutes = [A]=?

t = 50 minute

[A]=[A]_o\times e^{-kt}

[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}

[A] = 0.2251 M

The concentration of A after 50 minutes = 0.2251 M

The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M

0.6749 M is the concentration of B after 50 minutes.

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Neutral atoms can be turned into positively charged ions by removing one or more electrons.

Explanation:

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Gaseous ammonia chemically reacts with oxygen O2 gas to produce nitrogen monoxide gas and water vapor. Calculate the moles of am
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Answer:

1.7 moles of ammonia, NH₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH₃ + 5O₂ —> 4NO + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted to produce 4 moles of NO.

Finally, we shall determine the number of mole of ammonia, NH₃, needed to produce 1.7 moles of nitrogen monoxide, NO. This can be obtained as follow:

From the balanced equation above,

4 moles of NH₃ reacted to produce 4 moles of NO.

Therefore, 1.7 moles of NH₃ will also react to produce 1.7 moles of NO.

Thus, 1.7 moles of ammonia, NH₃, is required.

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A greenhouse is filled with air that cotains more carbon dixoide than normal air has. How might phothsyphness and plant growth b
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Answer:

See explanation

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If the atmosphere is rich in carbon dioxide such as  in a green house where air is filled with carbon dioxide, the rate of photosynthesis is increased.

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When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+
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Answer:

D) 65.7%

Explanation:

Based on the reaction:

2H2(g)+O2(g)⟶2H2O(l)

<em>2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.</em>

<em />

To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.

<em>Theoretical yield:</em>

Moles of 5.58g H₂:

5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂

As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:

2.768 moles H₂O ₓ (18.015g / mol) =

49.86g H₂O is theoretical yield

<em>Percent yield:</em>

Percent yield = Actual yield / Theoretical yield ₓ 100

32.8g H₂O / 49.86g ₓ 100 =

65.7% is percent yield of the reaction

<h3>D) 65.7% </h3>

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when positively charged particles were radiated onto a gold atom, most of the particles went straight past. what is most likely
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