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vladimir2022 [97]
3 years ago
10

You have a radioactive isotope with a mass of 400 grams. The sample has a half life of 5 days. How much is left after 25 days ha

ve passed?
Chemistry
1 answer:
Alexeev081 [22]3 years ago
4 0

Answer:

Amount left after 25 days = 12.5 g

Explanation:

Given data:

Mass of sample = 400 g

Half life of sample = 5 days

Mass left after 25 days = ?

Solution:

First of all we will calculate the number of half lives passes in given time period.

Number of half lives = Time elapsed / Half life

Number of half lives = 25 days/ 5 days

Number of half lives = 5

At time zero = 400 g

At 1st half life = 400 g/2 = 200 g

At 2nd half life = 200 g/2 = 100 g

At 3rd half life = 100 g/2 = 50 g

At 4th half life = 50 g/2 = 25 g

At 5th half life = 25 g/2 = 12.5 g

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A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
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