Answer:
the answer to that question is d
Answer:
The answer to your question is: kc = 6.48
Explanation:
Data
Given Molecular weight
CaO = 44.6 g 56 g
CO₂ = 26 g 44 g
CaCO₃ = 42.3 g 100 g
Find moles
CaO 56 g ---------------- 1 mol
44.6 g -------------- x
x = (44.6 x 1) / 56 = 0.8 mol
CO₂ 44 g ----------------- 1 mol
26 g ---------------- x
x = (26 x 1 ) / 44 = 0.6 moles
CaCO₃ 100 g --------------- 1 mol
42.3g -------------- x
x = (42.3 x 1) / 100 = 0.423 moles
Concentrations
CaO = 0.8 / 6.5 = 0.12 M
CO₂ = 0.6 / 6.5 = 0.09 M
CaCO₃ = 0.423 / 6.5 = 0.07 M
Equilibrium constant = ![\frac{[products]}{[reactants]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bproducts%5D%7D%7B%5Breactants%5D%7D)
Kc = [0.07] / [[0.12][0.09]
Kc = 0.07 / 0.0108
kc = 6.48
Answer:
I think it's B but I could be wrong so really sorry if I am
The Answer Is B Because You Have To Divide The 9
Answer:
16974J of energy are required
Explanation:
The energy required is:
* The energy to heat solid water from -15°C to 0°C using:
q = m*S*ΔT
* The energy to convert the solid water to liquid water:
q = dH*m
* The energy required to increase the temperature of liquid water from 0°C to 75°C
q = m*S*ΔT
The first energy is:
q = m*S*ΔT
<em>m = Mass water = 25g</em>
<em>S is specific heat of ice = 2.03J/g°C</em>
<em>ΔT is change in temperature = 0°C - (-15°C) = 15°C</em>
q = 25g*2.03J/g°C*15°C
q = 761.3J
The second energy is:
q = dH*m
<em>m = Mass water = 25g</em>
<em>dH is heat of fusion of water = 80cal/g</em>
q = 80cal/g*25g
q = 2000cal * (4.184J/1cal) = 8368J
The third energy is:
q = m*S*ΔT
<em>m = Mass water = 25g</em>
<em>S is specific heat of water= 4.184J/g°C</em>
<em>ΔT is change in temperature = 75°C-0°C = 75°C</em>
q = 25g*4.184J/g°C*75°C
q = 7845J
The energy is: 7845J + 8368J + 761J =
16974J of energy are required
