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3 years ago

Calculate the number of grams of HNO3 which must be added to 31.5 g of H20 to prepare a 0.950 m solution.

Chemistry

1 answer:

IrinaK [193]3 years ago

3 0

Answer:

1.89g of HNO3

Explanation:

Data obtained from the question include:

Mass of water = 31.5 g

Molality of HNO3 = 0.950 m

Mass of HNO3 =.?

Next, we shall determine the number of mole of HNO3 in the solution.

Mass of water = 31.5 g = 31.5/1000 = 0.0315 Kg

Molality of HNO3 = 0.950 m

Number of mole of HNO3 =..?

Molality is simply defined as the mole of solute per unit kilogram of solvent (water) i.e

Molality = mole /kg of water

With the above formula, the mole of HNO3 can be obtained as follow:

Molality = mole /kg of water

0.950 = mole of HNO3 /0.0315

Cross multiply

Mole of HNO3 = 0.950 x 0.0315

Mole of HNO3 = 0.03 mole

Finally, we shall convert 0.03 mole of HNO3 to grams. This is illustrated below:

Molar mass of HNO3 = 1 + 14 + (16x3) = 63g/mol

Mole of HNO3 = 0.03 mole

Mass of HNO3 =..?

Mole = mass /molar mass

0.03 = mass of HNO3 /63

Cross multiply

Mass of HNO3 = 0.03 x 63

Mass of HNO3 = 1.89g

Therefore, 1.89g of HNO3 is needed to prepare the solution.

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3 years ago

Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho

AleksandrR [38]

The answer is (3) 3 × 12.4 hours

To calculate this, we will use two equations:
$(1/2) ^{n} =x$
$t_{1/2} = \frac{t}{n}$
where:
n - number of half-lives
x - remained amount of the sample, in decimals
 $t_{1/2}$ - half-life length
t - total time elapsed.

First, we have to calculate x and n. x is remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams × 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
 $(1/2) ^{n} =x$
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$n*log(0.5)=log(0.125)$
$n= \frac{log(0.5)}{log(0.125)}$
$n=3$

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
 $t_{1/2} = 12.4$
 $t_{1/2} *n = t$
 $t= 12.4*3$

Therefore, it must elapse 3 × 12.4 hours before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope

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3 years ago

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