Answer:
19.8 kg of C₂H₂ is needed
Explanation:
We solve this by a rule of three:
If 1251 kJ of heat are relased in the combustion of 1 mol of acetylene
95.5×10⁴ kJ of heat may be released by the combustion of
(95.5×10⁴ kJ . 1) /1251kJ = 763.4 moles of C₂H₂
Let's convert the moles to mass → 763.4 mol . 26 g/1 mol = 19848 g
If we convert the mass from g to kg → 19848 g . 1kg / 1000g = 19.8 kg
Answer:
what are we doing there???
bye, have a nice day!
Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
They connect at the joint
