Solution is here,
for initial case,
temperature(T1)=70°C=70+ 273=343K
vloume( V1) =45 L
for final case,
temperature( T2)=?
volume(V2)= 91.3 L
at constant pressure,
V1/V2 = T1/T2
or, 45/91.3 = 343/ T2
or, T2= (343×91.3)/45
or, T2=695.9 K = (695.9-273)°C=422.9°C
Answer : Right
Explanation : The direction of reaction tends to proceed on right side under standard conditions; If the change in standard free energy ΔG for a particular reaction is negative. Also if the elements in their most stable forms as they exist under standard conditions. Then ΔG determines the direction and extent of chemical change. But under standard conditions the direction of the reaction will be to right.
Answer : The mass of sulfuric acid needed is 
.
Solution : Given,
pH = 8.94
Volume of solution = 380 ml = 
Molar mass of sulfuric acid = 98.079 g/mole
As we know,
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
![5.06=-log[OH^-]](https://tex.z-dn.net/?f=5.06%3D-log%5BOH%5E-%5D)
![[OH^-]=0.00000871=8.71\times 10^{-6}mole/L](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.00000871%3D8.71%5Ctimes%2010%5E%7B-6%7Dmole%2FL)
Now we have to calculate the moles of 
.
Formula used : 
![\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%5Ctext%7B%20Concentration%20of%20%7D%5BOH%5E-%5D%5Ctimes%20Volume%5C%5C%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%288.71%5Ctimes%2010%5E%7B-6%7Dmole%2FL%29%5Ctimes%20%28380%5Ctimes%2010%5E%7B-3%7DL%29%3D3309.8%5Ctimes%2010%5E%7B-9%7Dmoles)
For neutralization, equal number of moles of 
ions will neutralize same number of ions.
![\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7D%5BOH%5E-%5D%3D%5Ctext%7B%20Moles%20of%20%7D%5BH%5E%2B%5D%3D3309.8%5Ctimes%2010%5E%7B-9%7Dmoles)
As, 
From this reaction, we conclude that
2 moles of ion is given by the 1 mole of 
moles of ion is given by 
moles of
Now we have to calculate the mass of sulfuric acid.
Mass of sulfuric acid = Moles of × Molar mass of sulfuric acid
Mass of sulfuric acid = 
Therefore, the mass of sulfuric acid needed is .
Take 46g / 46 g/mol = 1 mol NO2 produced
1 mol * 1/2 = 0.5 mol O2 needed (1/2 ratio between NO2 and O2)
