Answer:
Player B
Explanation:
I just did it and I got it right :)
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
Answer:
Option B. 10
Explanation:
If 1 mol of butanol contains 6×10²⁴ atoms of H, let's calculate the amount of H.
(number of atoms / NA)
6.02 x 10²³ atoms ___ 1 mol
6×10²⁴ atoms will occupy (6×10²⁴ / NA) = 9.96 moles
H, has 10 moles in the butano formula.
Answer:
0.13 M
Explanation:
The reaction equation is;
NaOH(aq) + KHC8H4O4(aq) ------> KNaC8H4O4(aq) + H2O(l)
Molar mass of KHP = 204.22 g/mol
Amount of KHP= mass/ molar mass = 0.3365 g/204.22 g/mol = 1.65 × 10^-3 moles
n= CV
Where;
C= concentration
V= volume in dm^3
n= number of moles
C= n/V = 1.65 × 10^-3 moles × 1000/250 = 6.6 × 10^-3 M
If 1 mole of KHP reacts with 1 mole of NaOH
1.65 × 10^-3 moles of KHP will react with 1.65 × 10^-3 moles of NaOH
From
n= CV
We have that only 12.44 ml of NaOH reacted
C= n/V = 1.65 × 10^-3 moles × 1000/12.44
C= 0.13 M
At the equivalence point, the KHP solution turned light pink.
5.5 moles
2.0=x/2.75
2.0*2.75= x/2.75 (2.75)
5.5=x
