Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
Ca + 2HCl = CaCl₂ + H₂
c=4.50 mol/l
v=2.20 l
n(HCl)=cv
m(Ca)/M(Ca)=n(HCl)/2
m(Ca)=M(Ca)cv/2
m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g
198 grams of Ca are needed
M1V1 = M2V2
.200 (.025) = 1.60 X 10 -2 (V2)
V2 = .315 L
1.60 x 10-2 M in 315 mL
Answer:
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Explanation:
The equilibrium constant indicates the % of the yield reaction and can shows where the reaction is going to be equilibrated.
It works with molar concentrations on the equilibrium and it does not consider the solids compounds
Kc also can be modified by the time of the reaction.
This reaction is:
CS₂ (g) + 4 H₂O(g) ⇌ CH₄ (g) + 2H₂S (g)
Kc = [H₂S]² . [CH₄] / [ H₂O]⁴ . [CS₂]
Answer:
In pair NaF and H2O both compounds exibit predominantly ionic bonding.