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Sindrei [870]
3 years ago
5

Tin-126 has a half life of 100,000 years. How many years does it take to

Chemistry
2 answers:
lana66690 [7]3 years ago
6 0

Answer:

300,000 years

Explanation:

100g=0 years

50g=100,00 years

25g=200,00 years

12.5g=300,000 years

Nostrana [21]3 years ago
3 0
300’000 yes I think she is correct
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Explanation:

Rivers may be contaminated by sewage.

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Lead(2) sulfate yields to lead(2)sulfite and oxygen
PolarNik [594]
A standardized test of what you want to do with your friends and you get to know what you think you want to be tre
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1,500 grams is equal to:<br> 1.5 kg<br> 1.5L<br> 1.5 cm<br> all of the above
Salsk061 [2.6K]
1kg is 1,000 grams plus .5 is 1.5 kg
8 0
3 years ago
A student isolated 15.6 g of product from a chemical reaction. She calculated that the reactions should have produced 18.4 g of
Snezhnost [94]

Answer:

The percent yield of this reaction is 84.8 % (option A is correct)

Explanation:

Step 1: Data given

The student isolated 15.6 grams of the product = the actual yield

She calculated the reaction should have produced 18.4 grams of product = the theoretical yield = 18.4 grams

Step 2: Calculate the percent yield

Percent yield = (actual yield / theoretical yield ) * 100 %

Percent yield = (15.6 grams / 18.4 grams ) * 100 %

Percent yield  = 84.8 %

The percent yield of this reaction is 84.8 % (option A is correct)

7 0
3 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
Read 2 more answers
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