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Bond [772]
3 years ago
11

Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (C2H5OH) using the standard enthalpies of fo

rmation of the reactants and products.C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O
Chemistry
1 answer:
PSYCHO15rus [73]3 years ago
5 0

Answer:

Explanation:

For the reaction

C2H5OH (l) + 3 O2(g) = 2CO2(g) + 3 H2O

We can calculate the  standard molar enthalpy of combustion using the standard enthalpies of formation of the species involved in the reaction according to Hess law:

ΔHºc =  2ΔHºf CO2 (g) + 3ΔHºfH2O(l)  - ( ΔHºf C2H5OH (l) - 3ΔHºfO2 (g) )

( we were not give the water state but we know we are at standard conditions so it is in its liquid state )

The ΔHºfs can be found in appropiate reference or texts.

ΔHºc =  2ΔHºf CO2 (g)+ 3ΔHºfH2O(l)  - (  ΔHºf C2H5OH (l) -+3ΔHºfO2 (g) )

= [ 2 ( -393.52 ) + 3 ( -285.83 ) ] - [(  -276.2 + 0 ) ] kJ

ΔHºc =  -1368.33 kJ

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How many grams are 3.01 × 1023 molecules of CuSO4?
zvonat [6]
Answer is: 79.8 grams of copper(II) sulfate.
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4 0
3 years ago
A box has a weight of 120 lbs and the bottom of the box is 12 in2 . What is the pressure the box exerts on the floor?
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7 0
3 years ago
How many moles of hydrogen gas can be produced if 0.78 moles of hydrochloric acid reacts with excess solid zinc according to the
White raven [17]

Answer:

The answer to your question is It will be formed 0.39 moles of H₂

Explanation:

Data

moles of H₂ = ?

moles of HCl = 0.78

moles of Zinc = excess

Balanced chemical reaction

              2 HCl  +  Zn  ⇒  1 H₂  +  ZnCl₂

Process

1.- Use proportions to solve this problem. Consider the coefficients of the balanced reaction.

               2 moles of HCl ---------------------- 1 mol of H₂

               0.78 moles of HCl -----------------  x

               x = (0.78 x 1) / 2

- Simplification

              x = 0.78 / 2

- Result

              x = 0.39 moles of H₂

3 0
3 years ago
When 15.3 g of sodium nitrate, NaNO3,was dissolved in water in a calorimeter, the temperature fell from 25.00oC to 21.56oC. If t
satela [25.4K]

Answer:

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

Explanation:

Heat lost by solution ad calorimeter = Q

Heat capacity of solution ad calorimeter = C = 1071 J/°C

Change in temperature = ΔT = 21.56°C - 25.00°C = -3.44°C

Q=C\times Delta T

Q=1071 J/^oC\times (-3.44^oC)=-3,684.24 J

Heat gained by sodium nitrate = -Q = -(-3,684.24 J)=3,684.24 J

Moles of sodium nitrate = \frac{15.3 g}{85 g/mol}=0.18 mol

When 0.18 mole of sodium nitrate was dissolved in water 3,684.24 joulesof heat was absorbed by it.

Then heat absorbed by 1 mole of sodium nitrate :

\frac{3,684.24 J}{0.18}=20,468 J=20.468 kJ

1 J = 0.001 kJ

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

8 0
3 years ago
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