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3 years ago

2. Chromium+oxygen - chromium(III) oxide balanced

Chemistry

1 answer:

sesenic [268]3 years ago

3 0

Answer:

4Cr + 3O2 —> 2Cr2O3

Explanation:

Information from the question include:

Chromium + oxygen -> chromium(III) oxide

From the word equation given above, the equation can be written as follow:

Cr + O2 —> Cr2O3

The equation can be balance by doing the following:

There are 2 atoms of O on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of Cr2O3 and 3 in front of O2 as shown below:

Cr + 3O2 —> 2Cr2O3

Now, we have 4 atoms of Cr on the right side and 1 atom on the left. It can be balance by putting 4 in front of Cr as shown below:

4Cr + 3O2 —> 2Cr2O3

Now the equation is balanced

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3 years ago

If 27.3% of a sample of silver-112 decays in 1.52 hours, what is the half-life (in hours to 3 decimal places)?

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<u>Answer:</u> The half life of the sample of silver-112 is 3.303 hours.

<u>Explanation:</u>

All radioactive decay processes undergoes first order reaction.

To calculate the rate constant for first order reaction, we use the integrated rate law equation for first order, which is:

$k=\frac{2.303}{t}\log \frac{[A_o]}{[A]}$

where,

k = rate constant = ?

t = time taken = 1.52 hrs

$[A_o]$ = Initial concentration of reactant = 100 g

[A] = Concentration of reactant left after time 't' = [100 - 27.3] = 72.7 g

Putting values in above equation, we get:

$k=\frac{2.303}{1.52hrs}\log \frac{100}{72.7}\\\\k= 0.2098hr^{-1}$

To calculate the half life period of first order reaction, we use the equation:

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

where,

$t_{1/2}$ = half life period of first order reaction = ?

k = rate constant = $0.2098hr^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{0.2098hr^{-1}}\\\\t_{1/2}=3.303hrs$

Hence, the half life of the sample of silver-112 is 3.303 hours.

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3 years ago

Two colorless solutions are mixed together bubbles form as the solution is stirred give to possible explanations for this result

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3 years ago

g A radioactive isotope of mercury, 197Hg, decays to gold, 197Au, with a disintegration constant of 0.0108hrs.-1. What % of the

weqwewe [10]

Answer:

7.49% of Mercury

Explanation:

Let N₀ represent the original amount.

Let N represent the amount after 10 days.

From the question given above, the following data were obtained:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Next, we shall convert 10 days to hours. This can be obtained as follow:

1 day = 24 h

Therefore,

10 days = 10 day × 24 h / 1 day

10 days = 240 h

Thus, 10 days is equivalent to 240 h.

Finally, we shall determine the percentage of Mercury remaining as follow:

Rate of disintegration (K) = 0.0108 h¯¹

Time (t) = 10 days

Percentage of Mercury remaining =?

Log (N₀/N) = kt /2.303

Log (N₀/N) = 0.0108 × 240 /2.303

Log (N₀/N) = 2.592 / 2.303

Log (N₀/N) = 1.1255

Take the anti log of 1.1255

N₀/N = anti log 1.1255

N₀/N = 13.3506

Invert the above expression

N/N₀ = 1/13.3506

N/N₀ = 0.0749

Multiply by 100 to express in percent.

N/N₀ = 0.0749 × 100

N/N₀ = 7.49%

Thus, 7.49% of Mercury will be remaining after 10 days

5 0

3 years ago

2. Chromium+oxygen - chromium(III) oxide balanced​

2. Chromium+oxygen - chromium(III) oxide balanced