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never [62]
3 years ago
15

What is the mass percent of hydrogen in hexanal?

Chemistry
2 answers:
AfilCa [17]3 years ago
8 0

Answer : The mass percentage of hydrogen in hexanal is, 12 %

Explanation: Given,

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

First we have to calculate the molar mass of hexanal.

Molar mass of hexanal (C_{6}H_{12}O) = 6(12)+12(1)+(16)=100g/mole

Now we have to calculate the mass of carbon.

As we now that there are 6 number of carbon atoms, 12 number of hydrogen atoms and 1 number of oxygen atoms.

The mass of hydrogen = 12\times 1=12g

Now we have to calculate the mass percentage of hydrogen in hexanal.

Formula used :

\%\text{ Mass percentage of hydrogen}=\frac{\text{Mass of hydrogen}}{\text{Mass of hexanal}}\times 100

Now put all the given values in this formula, we get:

\%\text{ Mass percentage of hydrogen}=\frac{12}{100}\times 100=12\%

Therefore, the mass percentage of hydrogen in hexanal is, 12 %

yKpoI14uk [10]3 years ago
5 0
12.1% is the mass percent of hydrogen in hexanal
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The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.

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Using the following thermochemical data, what is the change in enthalpy for the following reaction? Ca(OH)2(aq) + HCl(aq) CaCl2(
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Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?

CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l),  Δ<span>H = -186 kJ
</span>
CaO(s) + H2O(l) -----> Ca(OH)2(s), Δ<span>H = -65.1 kJ
</span>
1) Ca(OH)2 should be  reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s) 
we are going to take as 
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2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l),                    and ΔH = 65.1 kJ
<span><u>CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ</u>

</span>Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)

Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add  enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ
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