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Roman55 [17]
2 years ago
11

For which of the following equations is the change in enthalpy at 25 Celcius and 1 atm pressure equal to ΔHf of CH3OH (l)?A.) CH

3OH + 3/2 O2 ---> CO2 + 2H2O (l)B.) CH3OH + 3/2 O2 ---> CO2 + 2H2O (g)C.) 2CH3OH + 3O2 ---> 2CO2 + 4H2O (l)D.) C + 2H2 + 1/2 O2 ---> CH3OH
Chemistry
1 answer:
Dahasolnce [82]2 years ago
8 0

Answer:

D. C + 2H₂ + ¹/₂O₂ → CH₃OH

Explanation:

The standard enthalpy of formation -ΔHf- is defined as the enthalpy change for the formation of 1 mol of a compound from its component elements.

Component elements are any element in its standard state.

For CH₃OH(l) at 25°C and 1 atm of pressure, the component elements are C(s), H₂(g) and O₂(g)

Thus, the equation that represent the standard enthalpy of formation of CH₃OH(l) is:

D. C + 2H₂ + ¹/₂O₂ → CH₃OH

I hope it helps!

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The pressure of a fixed mass of gas is increased from 100 kPa to 600 kPa at a constant temperature. The new volume of the gas is
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8 0
2 years ago
V1T2 = V2T1 is an expression of who’s law
Andru [333]

<u>Answer:</u>

It is the expression of Charles' Law.

<u>Explanation:</u>

The given expression V1T2 = V2T1 is the formula for the Charles' Law.

A special case of an ideal gas is named as the Charles' Law. This law applies to ideal gases only which are at constant pressure.

According to this law, the volume of a fixed mass of a gas is directly proportional to its temperature and is given by:

V1T2 = V2T1

6 0
3 years ago
Read 2 more answers
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