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2 years ago

11

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1 answer:

Phoenix [80]2 years ago

5 0

Answer:

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Explanation:

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Which descriptions apply to emitted radiation? Check all that apply. measurement in Ci/Bq measurement in conventional rad or the

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A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157g of the compound produced 0.213g of CO2 and

vladimir2022 [97]

Answer:

$C_{7} H_{5}N_{3}O_{6}$

Explanation:

First reaction gives you the number of moles or the mass from Carbon and hydrogen

for carbon:

$0,213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} =0.005molC$

$0.213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} .\frac{12gC}{1molC} = 0.058gC\\$

Analogously for hydrogen:

0.0310g $H_{2}O$ have 0.0034gH or 0.0034mol of H

In the second reaction you can obtain the amount of nitrogen as a percentage and find the mass of N in the first sample.

$0.023gNH_{3} .\frac{1molNH_{3}}{17gNH_{3}} .\frac{1molN}{1molNH_{3}} .\frac{14gN}{1molN} \frac{100}{0.103gsample} =18.4%N$

now

$\frac{18.4gN}{100gsample} .0.157gsample=0.0289gN in the first reaction$

this is equivalet to 0.002mol of N

with this information you can find the mass of oxygen by matter conservation.

$gO=total mass-(gN+gC+gH)=0.157-(0.0289+0.058+0.0034)=0.0666gO$

this is equivalent to 0.004molO

finally you divide all moles obtained between the smaller number of mole (this is mol of H)

$C\frac{0.0048}{0.0034} H\frac{0.0034}{0.0034} N\frac{0.002}{0.0034} O\frac{0.004}{0.0034} =C_{1.4} HN_{0.6} O_{1.2}$

and you can multiply by 5 to obtain: $C_{7} H_{5}N_{3}O_{6}$

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