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:
0.1035 M
Explanation:
Considering:
Sodium chloride will furnish Sodium ions as:
Given :
For Sodium chloride :
Molarity = 0.288 M
Volume = 3.58 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 3.58×10⁻³ L
Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:
Moles of sodium ions by sodium chloride = 0.00103104 moles
Sodium sulfate will furnish Sodium ions as:
Given :
For Sodium sulfate :
Molarity = 0.001 M
Volume = 6.51 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 6.51 ×10⁻³ L
Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:
Moles of sodium ions by Sodium sulfate = 0.00001302 moles
Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles
Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L
Concentration of sodium ions is:
<u> The final concentration of sodium anion = 0.1035 M</u>