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FrozenT [24]
3 years ago
13

Combustion analysis of an unknown compound containing only carbon and hydrogen produced 0.2845 g of co2 and 0.1451 g of h2o. wha

t is the empirical formula of the compound?
Chemistry
1 answer:
Serjik [45]3 years ago
7 0
CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O

m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}

0.2845/{44.01x}=0.1451/{9.01y}

x/y=0.4=2:5

C₂H₅ is the empirical formula 
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8 0
3 years ago
Arrange the following substances in order of increasing solubility of water. C6H14, C6H13Br, C6H13OH, C6H12(OH)2.
Juli2301 [7.4K]

Answer:

C6H14 < C6H13Br  < C6H13OH < C6H12(OH)2

Explanation:

Hello,

In this case, since the solubility in water is related with the presence of polar bonds in the given molecules we can see that C6H12(OH)2 has the presence two O-H bonds which promote the highest solubility via hydrogen bonds as well as the C6H13OH but in a lower degree as only on O-H bond is present. Next since the bond C-Br in is slightly close to the polar bond C6H13Br rather than the C-C bonds only had by C6H14 we can infer that C6H13Br is more soluble in water than C6H14, therefore the required order is:

C6H14 < C6H13Br  < C6H13OH < C6H12(OH)2

Whereas C6H12(OH)2 is the most soluble and C6H14 the least soluble in water.

Best regards.

5 0
3 years ago
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