Solute is something that is being dissolved { ex : sugar , salt}
Solvent is something that has ability to dissolve things { ex : water}
False because the solvent is present in larger amounts...
The pH of the monoprotic weak acid is 2.79.
<h3>What are weak acids?</h3>
The weak acids are the acids that do not fully dissociate into ions in the solution. Strong acids fully dissociate into ions.
The chemical reaction is HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c (monoprotic acid) = 0.33 M.
Ka = 1.2·10⁻⁶
[A⁻] = [H⁺] = x
[HA] = 0.33 M - x
Ka = [A⁻]·[H⁺] / [HA]
2. 6 × 10⁻⁶ = x² / (0.33 M - x)
Solve quadratic equation: [H⁺] = 0.000524 M.
pH = -log[H⁺]
pH = -log(0.000524 M)
pH = 2.79
Thus, the pH of the monoprotic weak acid is 2.79
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Answer is: molarity of hydrofluoric solution is 0.09 M.
Chemical reaction: HF(aq) + KOH(aq) → KF(aq) + H₂O(l).
V(HF) = 30.0 mL.
c(KOH) = 0.122 M.
V(KOH) = 22.15 mL:
c(HF) = ?.
From chemical reaction: n(HF) : n(KOH) = 1 : 1.
n(HF) = n(KOH).
c(HF) · V(HF) = c(KOH) · V(KOH).
c(HF) = c(KOH) · V(KOH) ÷ V(HF).
c(HF) = 0.122 M · 22.15 mL ÷ 30 mL:
c(HF) = 0.09 M.
True? Not sure what the question is
42.34 g of water could be warmed from 21.4°C to 43.4°C by the pellet dropped inside it
Heat loss by the pellet is equal to the Heat gained by the water.
….(1)
where, 
is the heat gained by water
is the heat loss by pellet
= mCΔT
where m = mass of water
C = specific heat capacity of water = 4.184 J/g-°C
ΔT = Increase in temperature
ΔT for water = 43.4 - 21.4 = 22°C
= m × 4.184 × 22 …. (2)
Now
= 
×ΔT
where = Heat capacity of pellet = 56J/°C
Δ T for pellet = 43.4 - 113 =- 69.6°C
= 56 × -69.6 = -3897.6 J
From equation (1) and (2)
-m× 4.184 × 22 =-3897.6
m= 42.34 g
Hence, 42.34 g of water could be warmed from 21.4 degrees Celsius to 43.4 degrees Celsius by the pellet dropped inside it.
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