Answer:
The correct option is C
Explanation:
From the question we are told that
The reaction is
Generally 
Here 
is the change in enthalpy
is the change in the internal energy
is the difference between that number of moles of product and the number of moles of reactant
Looking at the reaction we can discover that the elements that was consumed and the element that was formed is 
and 
and this are both gases so the change would occur in the number of moles
So
![\Delta H = \Delta U + [16 -23]* RT](https://tex.z-dn.net/?f=%5CDelta%20%20H%20%20%3D%20%20%5CDelta%20%20U%20%2B%20%5B16%20-23%5D%2A%20%20RT)
The negative sign in the equation tell us that the enthalpy
would be less than the Internal energy 
Moles of CO₂ = mass / molecular weight
Moles of CO₂ = 4.4 / (12 + 16 x 2)
Moles of CO₂ = 0.1 mol
Each mole of gas occupies 22.4 L at STP. Therefore,
Moles of NH₃ = 5.6 / 22.4
Moles of NH₃ = 0.25 mol
Answer : The concentration of NaOH is, 0.336 M
Explanation:
To calculate the concentration of base, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
Thus, the concentration of NaOH is, 0.336 M
Answer:
6.68 X 10^-11
Explanation:
From the second Ka, you can calculate pKa = -log (Ka2) = 6.187
The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,
8.181 = (6.187 + pKa3) / 2
Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11
