Explanation:
here's the answer to your question
Answer:
correct option is (a)
The solution would be using this: C6H5COOH = H+ + C6H5COO Ka = 6.5 x 10^-5 = (H+)(C6H5COO-) over
(C6H5COOH)
Let X = moles per liter (H+) and also = moles per liter (C6H5COO-)
Ka = 6.5 x 10^-5 = (X)(X) over .350 molar = acid solution 6.5 x 10^-5 = X^2 over .350
X^2 = 6.5 x 10^-5 times .350 which = 2.275 x 10^-5
x = V2.275 x 10^-5
X = 1.5083 x 10^-5 moles per liter H+
pH = -log(H+) = -log 1.5083 x 10^-5 which
= 4.6215
Answer:
- <em>2NaCl → 2Na + Cl₂, ΔH = 822 kJ </em>
Explanation:
The chemical <em>equation</em> for the <em>formation of NaCl</em> is:
- Na + (1/2) Cl₂ → NaCl , ΔH = - 411 kJ
That equation means that 1 mole of NaCl is formed by the reaction of 1 mole of Na and 1/2 mole of Cl₂, with a release of energy of 411 kJ.
The <em>decomposition</em> of <em>NaCl</em> is the inverse of the <em>formation</em> reaction; thus, you swift products and reactants and inverse the sign of the <em>change in enthalpy:</em>
- NaCl → Na + 1/2 Cl₂, ΔH = 411 kJ
Since you want the decomposition of 2 moles you multiply the equation and the ΔH by 2:
- 2NaCl → 2Na + Cl₂, ΔH = 822 kJ ← answer
