12.2 C
It has 3 significant figures now.
Answer:
oxidation number is correct!! :)
Explanation:
<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:
Explanation:
formula of osmotic pressure is as follows
p= n RT
n is mole of solute per unit volume
If m be the grams of solute needed
m gram = m / 227.1 moles
m / 227.1 moles dissolved in .279 litres
n = m / (227.1 x .279 )
= m / 63.36
substituting the values in the osmotic pressure formula
5.14 = (m / 63.36) x .082 x 298
m / 63.36 = .21
m = 13.32 grams .
Answer:
C . 24 L
Explanation:
Given data:
Initial volume of gas = 20.0 L
Initial pressure of gas = 660 mmHg
Final volume = ?
Final pressure = 550 mmHg
Solution:
The given problem will be solved through the Boly's law,
"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
Now we will put the values in formula,
P₁V₁ = P₂V₂
660 mmHg × 20.0 L = 550 mmHg × V₂
V₂ = 13200 mmHg. L/ 550 mmHg
V₂ = 24 L
