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barxatty [35]
3 years ago
6

The heaviest known isotope of hydrogen is called tritium, . it decays by beta emission, and has a half-life of 12.3 years. what

fraction of a tritium sample will remain after 5.20 years?
Chemistry
1 answer:
viva [34]3 years ago
6 0

Answer:

= 3/4

Explanation:

Half life is the time taken by a radioactove element to decay by half of its original mass.

The half life of tritium is 12.3 years , therefore it would take 12.3 years for tritium to decay to a half its original mass.

Using the formula;

New mass = Initial mass × (1/2)^n ; where n is the number of half lives.

n = 5.2 years/12.3 years

  = 52/123

Therefore;

New mass = 1 × (1/2)^(52/123)

                  = 0.74599

                  = 0.75

The fraction of a tritium sample after 5.20 years will be approximately 3/4

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Answer:

10B has 18.9%

11B has 81.1%

Explanation:Please see attachment for explanation

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3 years ago
Which of the following is a reason to make an armature the parent of a creature
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Balance the following chemical equation by providing the correct coefficients fe+h2so4 fe(so4)3 + h2
Dmitriy789 [7]

Answer:

2Fe + 3H2SO4 + Fe2(SO4)3+ 3H2

Explanation:

1. Fe (SO4) 3 is an incorrectly written formula because iron is trivalent as we can see by this three ahead of SO4. SO4 is divalent always.

2. since (SO4) is 3, this three shows us that there must be 3 in the reactants as well.

so now there is 3H2SO4

3. Since we have added 3 to one hydrogen we must add another. So now it's 3H2

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3 0
2 years ago
Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

8 0
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