Answer:
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
Explanation:
...[1]
..[2]
..[3]
The standard enthalpy of formation of ethanoic acid :
..[4]
Using Hess's law to calculate :
2 × [1] + 2 × [2] - [3] = [4]
The standard enthalpy of formation of ethanoic acid is -484 kJ/mol.
Answer:
0.0042 M is the molarity of tartaric acid in this sample of wine.
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is tartaric acid
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
0.0042 M is the molarity of tartaric acid in this sample of wine.
What kind of question is that it’s Hot
Answer:
8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.
Explanation:
Moles of ethanol of ethanol = 0.200 mol
Heat required to heat 0.200 moles of ethanol = Q = 1.05 kJ
Enthalpy of vaporization of ethanol =
Heat required to vaporize 0.200 moles of ethanol = Q'
Total heat required to fore heating and the vaporizing :
= Q + Q' = 1.05 kJ + 7.72 kJ = 8.77 kJ
8.77 kilo Joules will be the total amount of heat required for both the heating and the vaporizing.
Answer:
When you freeze the already cracked glow stick the chemical reactions are slowed down. When you take it out of the freezer and it reaches room temperature the chemical reaction will go on as 'normal'. XD hope this helps
Explanation: