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3 years ago

A certain ore is 37.3% nickel by mass how many kilograms of this ore would you need to dig up to have 10.0g of nickel

1 answer:

7 0

If the grade of the ore is 37.3% nickel, then the unknown quantity to get 10 grams of nickel is 0.373 x = 10 grams or x = 10/0.373=26.8 grams or 0.0268 kg needed to dig up to recover the 10 grams of nickel. At this grade of ore, 1 kilogram would yield 373 grams of nickel.

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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)?

ICE Princess25 [194]

<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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