i dont have paper but heres where they should go
5.22*22^3 should be the answer
The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
Answer is: the equilibrium concentrations fluorine anion are 0.004 M and lead cation are 0.002 M.<span>
Chemical reaction: PbF</span>₂(aq) → Pb²⁺(aq) + 2F⁻(aq).<span>
Ksp = 3,2·10</span>⁻⁸.
[Pb²⁺] = x.
[F⁻] = 2[Pb²⁺] = 2x<span>
Ksp = [Pb²</span>⁺] ·
[F⁻]².
Ksp = x · 4x².
3,2·10⁻⁸ = 4x³.
x = ∛3,2·10⁻⁸ ÷ 4.
x = [Pb²⁺] = 0,002M = 2·10⁻³ M.
[F⁻] = 2 · 0,002M = 0,004 M = 4·10⁻³ M.
Answer:
Heat transfer during melting of ice plays greater role in cooling of liquid water.
Explanation:
Temperature of ice = -10 °c
Temperature of water = 0 °c
When ice cube is dipped in to the water.the heat transfer
Q = m c ΔT
⇒ Q = 1 × 2.01 × 10
⇒ Q = 20.1 KJ
Heat transfer during melting of ice
= latent heat of ice
Latent heat of ice = 334 KJ
⇒
= 334 KJ
Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.
Thus heat transfer during melting of ice plays greater role in cooling of liquid water.