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scoray [572]
3 years ago
10

What is the total number of years that must pass before only 25.00 grams of an original 100.0-gram sample of C-14 remains unchan

ged?
(1) 2,865 y (3) 11,460 y
(2) 5,730 y (4) 17,190 y
Chemistry
2 answers:
Karolina [17]3 years ago
8 0
The answer is (3) 11,460 y. For this problem, you must know that the half life of C-14 is 5740 y and that nuclear decay processes are first order reactions (which means that the half life remains constant). For 25.00g to be left of a 100.0g sample, two half life must have elapsed (100*0.5*0.5=25). Each half life is 5730 y, so the total time elapsed is 5730 * 2 = 11,460 y.
REY [17]3 years ago
6 0

Answer:

The answer is 3; 11,460

Explanation:

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Answer:

hope this helps :)

Explanation:

for the first one, you can look at the periodic table and look at the atomic number and it will show you how many protons there are giving you the answer because protons and electrons are equal in a pure element

a- carbon

b- neon

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5 0
2 years ago
Read 2 more answers
Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1
givi [52]

According to Hasselbach-Henderson equation:

pH=pK_{a}+log\frac{[A^{-}]}{[HA]}

Here, [A^{-}] is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is CH_{3}COOK and acid is CH_{3}COOH thus, Hasselbach-Henderson equation will be as follows:

pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{2}{1}=5.06

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{3}=4.28

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{5}{1}=5.45

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{1}=4.76

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{10}=3.76

Therefore, pH of solution is 3.76.

4 0
3 years ago
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