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algol13
3 years ago
7

Consider the following thermochemical reaction for kerosene:

Chemistry
2 answers:
ch4aika [34]3 years ago
7 0

We have to solve this question using the stoichiometry of the reaction:

The equation of the reaction is;

2 C12H26(l) + 37 O2(g) -----> 24 CO2(g) + 26 H2O(l) + 15,026 kJ

According to the question;

Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles

From the  stoichiometry of the reaction:

Since;

24 moles of CO2 released 15,026 KJ

0.48 moles of CO2 will release 0.48 * 15,026/24

= 301 KJ of heat.

brainly.com/question/6901180

julsineya [31]3 years ago
3 0

Answer:

2 C12H26(l) + 37 O2(g) ----- > 24 CO2(g) + 26 H2O(l) + 15,026 kJ2C12H26(l)+37O2(g)−−−−−>24CO2(g)+26H2O(l)+15,026kJ

According to the question;

Number of moles of CO2 released = 21.3 g/44 g/mol = 0.48 moles

Since;

24 moles of CO2 released 15,026 KJ

0.48 moles of CO2 will release 0.48 * 15,026/24

= 301 KJ of heat.

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Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass
denis23 [38]

Answer:

607 ppm

Explanation:

In this case we can start with the <u>ppm formula</u>:

ppm=\frac{mg~of~solute}{Litters~of~solution}

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In other words:

0.0320~M=\frac{mol}{1~L}

mol=0.032~M*1~L=0.032~mol

1~L~of~Solution=0.0320~mol~of~solute

If we use the <u>atomic mass</u> of F  (19 g/mol) we can convert from mol to g:

0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g

Now we can <u>convert from g to mg</u> (1 g= 1000 mg), so:

0.607~g\frac{1000~mg}{1~g}=607~mg

Finally we can <u>divide by 1 L</u> to find the ppm:

ppm=\frac{607~mg}{1~L}=~607~ppm

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Answer:

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4 0
3 years ago
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