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maw [93]
3 years ago
8

Which of the following is made up of only one type of atom? 1) NO 2) He 3)CO 4)MgS

Chemistry
1 answer:
olga55 [171]3 years ago
5 0
2)He is made up of one atom
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The balanced equation for combustion in an acetylene torch is shown below: 2C2H2 + 5O2 → 4CO2 + 2H2O The acetylene tank contains
andriy [413]

Answer: 67.2 moles

Explanation: 2C_2H5+5O_2\rightarrow 4CO_2+2H_2O

According to the given balanced equation, 2 moles of acetylene C_2H_2 combine with 5 moles of oxygen O_2  to produce 4 moles of carbon dioxide CO_2.

Thus if 2 moles of acetylene C_2H_2 combine with = 5 moles of oxygen O_2

35 moles of acetylene C_2H_2 combine with=\frac{5}{2}\times {35}=87.5 moles of oxygen O_2

But as only 84 moles of oxygen are available, acetylene is not a limiting reagent.

5 moles of oxygen O_2 reacts with = 2 moles of acetylene C_2H_2

84 moles of oxygen O_2 reacts with=\frac{2}{5}\times {84}=33.6 moles of acetylene C_2H_2

Thus Oxygen is the limiting reagent as it limits the formation of products. Acetylene is excess reagent as it is present in excess.

2 moles of acetylene C_2H_2  produce= 4 moles of carbon dioxide CO_2.

33.6 moles of acetylene C_2H_2 produce=\frac{4}{2}\times {33.6}=67.2 moles of of carbon dioxide CO_2.


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3 years ago
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larisa [96]
A. is the answer
if u have any doubts ask me
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Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.
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The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
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0.78  is what the verified answer says 
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