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densk [106]
3 years ago
13

1. Calculate the empirical formula of a compound that contains 2% hydrogen, 74 % carbon and 24% oxygen.

Chemistry
1 answer:
Ludmilka [50]3 years ago
5 0

empirical formula = when the percentage compositions by mass of elements in a compound are divided by their relative atomic masses.

from the obtained amount the smallest amount will divide all.

the answers can then be rounded off if in decimals or they can be multiplied by 2 if they have .5 decimals e.g 2.5 , 7.5 , etc.

solution: HYDROGEN    CARBON      OXYGEN

                2/1                      74/12               24/16

                  2                         6.167                 1.5

                  2 / 1.5                  6.167 / 1.5           1.5 / 1.5

                  1.333                        4.111                   1

                      1                              4                        1


answer :     C4OH

HOPE IT HELPED !!!

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Answer:

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Explanation:

a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.

The area of the first peak corresponding to Chlorohexane is 32434 units.

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Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:

1.69 nmols of Chlorohexane gives 32434 units

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Incomplete question, it is lacking the data it makes reference. The missing data from Chegg is:

                              2 SO3(g)   →          2 SO2(g) + O2(g)

ΔHf° (kJ mol-1)  -395.7                        -296.8

S° (J K-1 mol-1)  256.8                         248.2              205.1

ΔH° =  kJ

S° =  J K⁻¹

Explanation:

The method to solve this problem calls for the use of the Gibbs standard free energy change:

ΔG = ΔrxnH - TΔSrxn

We know a reaction is spontaneous when ΔG is < 0, so to answer this question we need to solve for the temperature, T, at which ΔG becomes negative.

Now as mentioned in the hint, we need to determine  ΔrxnH and ΔSrxn, which are given by

ΔrxnH = ∑ ν x ΔfHº products - ∑ ν x ΔfHº reactants

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For ΔS we have likewise

ΔrxnS =  ∑ ν x ΔSº products - ∑ ν x ΔSº reactants

Thus,

ΔrxnH(kJmol⁻¹) =  2 x (-296.8) - 2 x ( -395.7 ) = 197.8 kJ

ΔrxnS ( JK⁻¹) = 2 x 248.2 + 205.1 - 2 x 256.8 = 187.9 JK⁻¹ = 0.1879 kJK⁻¹

So ΔG kJ =  197.8 - T(0.1879)

and the reaction will become spontaneous when the term  T(0.1879)  becomes greater that 197.8,

0 = 197.8 - 0.1879 T  ⇒ T = 1052 K

so the reaction is spontaneous at temperatures greater than 1052 K (780 ºC)

4 0
3 years ago
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