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katovenus [111]
3 years ago
6

The half-life of carbon-14 is 5730 years. How many grams of carbon-14 remain after 28,650 years if the original amount was 28.0

g?
A.5.60 g B.2.80 g C.0.875 g D.14.0 g
Chemistry
1 answer:
alexdok [17]3 years ago
5 0
TLDR: the answer is C. 0.875.

The basis of this problem is testing your fluency with half-life radioactive decay. The half-life of any radioactive substance represents the amount of time required for 1/2 of the amount of the substance to decay. So, if you had a 100 g sample of Carbon-14, after 5,730 years, only 50 g of Carbon-14 would remain; the other 50 g of substance would have decayed into another substance (Nitrogen-14). If another 5,730 years passes, only 25 g would remain, as half of that 50 g radioactive substance would have decayed.

In the problem, 28 g of Carbon-14 decays as 28,650 years passes, which, when divided by 5,730, yields a value of 5. This number represents the number of half lives that passes during this time. So, over the course of this time, the amount of radioactive substance is cut in half five times.

28 g - 14 g (1 HL) - 7 g (2 HL) - 3.5 (3 HL) - 1.75 g (4 HL) - 0.875 g (5 HL)

After 5 half lives, only 0.875 g of the original 28 g remains.
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Answer:

49.45~%

Explanation:

In this case, we have to start with the <u>chemical reaction</u>:

C_6H_1_2O~->~C_6H_1_0~+~H_2O

So, if we start with <u>10 mol of cyclohexanol</u> (C_6H_1_2O) we will obtain 10 mol of cyclohexanol (C_6H_1_0). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

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With this value we can calculate the grams:

10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0

Now, we have as a product 500 mL of C_6H_1_0. If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

500~mL\frac{0.811~g}{1~mL}=405.5~g

Finally, with these values we can calculate the <u>yield</u>:

%~=~\frac{405.5}{820}x100~=~49.45%%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

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