The final temperature of the mixture : 21.1° C
<h3>Further explanation </h3>
The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released
Q in(gained) = Q out(lost)
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
Q ethanol=Q water
mass ethanol=
mass water =
then the heat transfer :
Answer:
P₂ = 0.09 atm
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 0.225 L
Initial pressure = 338 mmHg (338/760 =0.445 atm)
Initial temperature = 72 °C (72 +273 = 345 K)
Final temperature = -15°C (-15+273 = 258 K)
Final volume = 1.50 L
Final pressure = ?
Solution:
P₁V₁/T₁ = P₂V₂/T₂
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L
P₂ = 25.83 atm .L. K / 293 K . L
P₂ = 0.09 atm
Ocean water freezes just like freshwater, but at Lower temperature. Fresh water freezes At 32°F but see water freezes at about 28.4°F because of the salt in it it can be melted down to use as drinking water
When we have:
Zn(OH)2 → Zn2+ 2OH- with Ksp = 3 x 10 ^-16
and:
Zn2+ + 4OH- → Zn(OH)4 2- with Kf = 2 x 10^15
by mixing those equations together:
Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6
by using ICE table:
Zn(OH)2 + 2OH- → Zn(OH)4 2-
initial 2m 0
change -2X +X
Equ 2-2X X
when we assume that the solubility is X
and when K = [Zn(OH)4 2-] / [OH-]^2
0.6 = X / (2-2X)^2 by solving this equation for X
∴ X = 0.53 m
∴ the solubility of Zn(OH)2 = 0.53 M
