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goldfiish [28.3K]
3 years ago
11

Using the data below, calculate the enthalpy for the combustion of C to CO

Chemistry
1 answer:
Leona [35]3 years ago
6 0

Answer:

ΔH3 = -110.5 kJ.

Explanation:

Hello!

In this case, by using the Hess Law, we can manipulate the given equation to obtain the combustion of C to CO as shown below:

C(s) + 1/2O2(g) --> CO(g)

Thus, by letting the first reaction to be unchanged:

C(s) + O2(g)--> CO2 (g) ; ΔH1 = -393.5 kJ

And the second one inverted:

CO2(g) --> CO(g) + 1/2O2(g) ; ΔH2= 283.0kJ

If we add them, we obtain:

C(s) + O2(g) + CO2(g) --> CO(g) + CO2 (g) + 1/2O2(g)

Whereas CO2 can be cancelled out and O2 subtracted:

C(s) + 1/2O2(g)  --> CO(g)

Therefore, the required enthalpy of reaction is:

ΔH3 = -393.5 kJ + 283.0kJ

ΔH3 = -110.5 kJ

Best regards!

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3 years ago
What is the difference between cellular respiration and photosynthesis
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3 0
3 years ago
How many moles of O₂ are needed to react completely with 35.0 mol of FeCl₃? *
densk [106]

Answer:

26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃

Explanation:

To determine the number of moles of O₂ that are needed to react completely with 35.0 mol of FeCl₃, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction),  and rule of three as follows: if 4 moles of FeCl₃ react with 3 moles of O₂, 35 moles of FeCl₃ with how many moles of O₂ will it react?

molesofO_{2} =\frac{35 moles of FeCl_{3}*3 moles of O_{2}  }{4 moles of FeCl_{3}}

moles of O₂= 26.25 ≅ 26.3

<u><em>26.3 moles of O₂ are needed to react completely with 35.0 mol of FeCl₃</em></u>

7 0
3 years ago
A sample of argon gas has a volume of 73 mL at a pressure of 1.20 atm and a temperature of 112 degrees Celsius. What is the fina
NeX [460]

Answer:

V₂ → 106.6 mL

Explanation:

We apply the Ideal Gases Law to solve the problem. For the two situations:

P . V = n . R . T

Moles are still the same so → P. V / R. T = n

As R is a constant, the formula to solve this is: P . V / T

P₁ . V₁ / T₁ = P₂ .V₂ / T₂   Let's replace data:

(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C

((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂

58.66 mL.atm = 0.55 atm . V₂

58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL

3 0
4 years ago
Which instrument is most often used to measure acid volume before a titration begins?
Darya [45]

<span>Good Morning!
 
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4 0
4 years ago
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