<span>Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN</span>⁻(aq)
+ Na⁺<span>(aq).
Chemical reaction 2: CN</span>⁻ +
H₂O(l) ⇄ HCN(aq) + OH⁻<span>(aq).
c(NaCN) = c(CN</span>⁻<span>)
= 0.021 M.
Ka(HCN) = 4.9·10</span>⁻¹⁰<span>.
Kb(CN</span>⁻) = 10⁻¹⁴ ÷
4.9·10⁻¹⁰ = 2.04·10⁻⁵<span>.
Kb = [HCN] · [OH</span>⁻]
/ [CN⁻<span>].
[HCN] · [OH</span>⁻<span>] =
x.
[CN</span>⁻<span>] = 0.021 M - x..
2.04·10</span>⁻⁵<span> = x² / (0.021 M
- x).
Solve quadratic equation: x = [OH</span>⁻<span>] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.</span>
Answer:- There are 
moles.
Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is 
and it's molar mass is 294.31 grams per mole.
mg are converted to grams and then the grams are converted to moles as:
= moles of aspartame
So, there would be moles of aspartame in 1.00 mg of it.
Answer:The Peruvian west coast bordering the Pacific Ocean is a long desert strip that stretches from the departments of Tumbes in the north bordering Ecuador, to Tacna in the south bordering Chile for a total length of 1,555 miles or 2,500 km. The average temperature from December to April is 25 to 28C or 53 to 59F and the average temperature from May to November is 12 to 15C or 53 to 59F.
The coast covers about 10% of the territory but is home to more than 50% of the population. Large cities such as Lima, the capital, Trujillo, Chiclayo and Tacna are located in the coast. Cities along the coast are the center of economic activity attracting people from the interior of the country looking for economic prosperity.
idk hope this helps !!!!
PH=-log[H⁺]
pH=-log(1.87×10⁻¹³)
pH=12.72
I hope this helps. Let me know if anything is unclear.
Solution:
At the equivalence point, moles NaOH = moles benzoic acid
HA + NaOH ==> NaA + H2O where HA is benzoic acid
At the equivalence point, all the benzoic acid ==> sodium benzoate
A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)
Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11
Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150
x^2 = 3.33x10^-12
x = 1.8x10^-6 = [OH-]
pOH = -log [OH-] = 5.74
pH = 14 - pOH = 8.26
