Answer:
I have solution to your question
Answer:
3.72L
Explanation:
Given parameters:
Initial volume V₁ = 3.9L
Condition = STP
Final temperature T₂ = -550°C
Final pressure P₂ = 880mmHg
Unknown:
Final volume V₂ = ?
Solution.
At standard temperature and pressure(STP), the:
Pressure = 1atm = 760mmHg
Temperature = 273K
Therefore, P₁ = 760mmHg
T₁ = 273K
The general gas law, is best to solve this problem. It is mathematically given as:
Let us take the units to the appropriate one;
-550°C = 273 + (-550) = -277K
Input the variables;
V₂ = 3.72L
Answer:
First, let's express pressure P in Pa and volume V in m3:
⇒P=175 kPa
⇒P=1.75×105 Pa
and
⇒V=275 ml
⇒V=2.75×10−4 m3
Then, let's solve the ideal gas law PV=nRT for temperature T:
⇒PV=nRT
⇒T=PVnR
Substituting the appropriate values into the equation:
⇒T=1.75×105×2.75×10−40.80×8.314 K
⇒T=48.1256.6512 K
∴T≈7.24 K
Therefore, the temperature is around 7.24 K.
A metal it produces carbon dioxide and water
The standard enthalpy of formation of ethane is determined as -192.5 kJ/mol.
<h3>
Reaction for the formation of ethane</h3>
The reaction for the formation of ethane is given as;
2C + 3H₂ → C₂H₆
Where;
- C is carbon (graphite)
- H is hydrogen
- C₂H₆ is ethane
<h3>Standard enthalpy of formation of ethane</h3>
The standard enthalpy of formation of ethane is calculated as follows;
ΔH = ΔH₁ + ΔH₂ + ΔH₃
where;
- ΔH₁ is heat of formation of graphite = -286 kJ/mol
- ΔH₂ is heat of formation of hydrogen = -393.5 kJ/mol
- ΔH₃ is heat of formation of ethane = 1560 kJ/mol
2C + 3H₂ → C₂H₆
ΔH = 2(-286) + 3(-393.5) + 1(1560)
ΔH = -572 - 1180.5 + 1560
ΔH = -192.5 kJ/mol
Learn more about standard enthalpy here: brainly.com/question/14047927
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