We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.
The final temperature is: 75.11 °C.
The work done at constant pressure, W=nR(T₂-T₁)
n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).
W=2.4 × 10³ Joule (Given)
From the expression,
(T₂-T₁)=
(T₂-T₁)= 
(T₂-T₁)= 48.11
T₂=300+48.11=348.11 K= 75.11 °C
Final temperature is 75.11 °C.
Answer:
Empirical formula of compound is C₄H₈O
Explanation:
Given data:
Mass of compound = 5.60 g
Mass of CO₂ = 13.7 g
Mass of H₂O = 5.60 g
Empirical formula of compound = ?
Solution:
Percentage of C:
13.7 g/5.60 g × 12/44× 100
2.45×0.273× 100 = 66.9%
Percentage of H:
5.60 g/ 5.60 g × 2.016/18 × 100
11.2%
Percentage of O:
(66.9% + 11.2%) - 100 = 21.9%
Grams atom of C , H, O
66.9/12 = 5.6
11.2 / 1.008 = 11.11
21.9 / 16 = 1.4
Atomic ratio:
C : H : O
5.6/1.4 : 11.11/1.4 : 1.4/1.4
4 : 8 : 1
Empirical formula:
C₄H₈O
The oxidation state of a substance is the electric charge it is exhibiting in a given state. This may be determined by looking at the oxidation states of accompanying atoms as well as the charge on the complete molecule.
In this case:
Molecular charge: 0
Oxidation state of oxygen: -2 (it is a group 6 element)
Thus,
S + 3 * -2 = 0
S = 6
Sulfur is exhibiting an oxidation state of +6.
