The HCl added = 1.25 moles
and the moles of Na2HPO4 = 1 mole
Now when acid is added in the given solution of Na2HPO4
One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4
Na2HPO4 + H+ ---> NaH2PO4
Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution
So this will result into formation of a buffer of phosphoric acid and NaH2PO4
NaH2PO4 + H+ ---> H3PO4
pKa of H3PO4 = 2.1
so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58
so the pH will be in between 2.1 to 7.2
Answer:
126 m
Explanation:
Distance=time*speed➡42*3=126m.
Answer:
heat energy
Explanation:
Chemical reactions often involve changes in energy due to the breaking and formation of bonds. Reactions in which energy is released are exothermic reactions, while those that take in heat energy are endothermic.
E. Traits are inherited when genes are passed from parents to offspring
Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2
when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
= 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value)
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.
c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.