If you mark those points on the phase diagram it appears to be moving from vapor to liquid. If you can’t see how I determined that please let me know and I can further explain.
Vapor to liquid is the same as saying gas to liquid, which I believe to be the correct answer.
Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present
The reaction will be
C3H8(g) + 5O2(g) ----> 3CO2(g) + 4H2O
so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2
now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O
Hence answer is
molecules of CO2 formed = 6
Molecules of H2O formed = 8
molecules of C3H8 left = 1
molecules of O2 left = 0
Answer:
The value of equilibrium constant is 29.45.
Explanation:
Moles of hydrogen gas = 2.00 mol
Concentration of hydrogen gas =![[H_2]= \frac{2.00 mol}{1.00 L}=2.00 M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%20%5Cfrac%7B2.00%20mol%7D%7B1.00%20L%7D%3D2.00%20M)
Moles of iodine gas = 1.00 mol
Concentration of iodine gas =![[I_2]= \frac{I.00 mol}{1.00 L}=1.00 M](https://tex.z-dn.net/?f=%5BI_2%5D%3D%20%5Cfrac%7BI.00%20mol%7D%7B1.00%20L%7D%3D1.00%20M)
![H2(g) + I2(g)\rightleftharpoons 2 HI(g)](https://tex.z-dn.net/?f=H2%28g%29%20%2B%20I2%28g%29%5Crightleftharpoons%202%20HI%28g%29)
initially
2.00 M 1.00 M 1.00 M
At equilibrium:
(2.00-x/2) (1.00-x/2) x
Moles of HI at equilibrium = 1.80 M
Concentration of HI at equilibrium =![[HI]=\frac{1.80 mol}{1.00L} = 1.80M= x](https://tex.z-dn.net/?f=%5BHI%5D%3D%5Cfrac%7B1.80%20mol%7D%7B1.00L%7D%20%3D%201.80M%3D%20x)
The expression of an equilibrium constant is given by ;
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
![K_c=\frac{x^2}{(2.00-\frac{x}{2})(1.00-\frac{x}{2})}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7Bx%5E2%7D%7B%282.00-%5Cfrac%7Bx%7D%7B2%7D%29%281.00-%5Cfrac%7Bx%7D%7B2%7D%29%7D)
Putting x equal to 1.80 M.
![K_c=\frac{(1.80)^2}{(2.00-\frac{1.80}{2})(1.00-\frac{1.80}{2})}=29.45](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%281.80%29%5E2%7D%7B%282.00-%5Cfrac%7B1.80%7D%7B2%7D%29%281.00-%5Cfrac%7B1.80%7D%7B2%7D%29%7D%3D29.45)
The value of equilibrium constant is 29.45.
Answer:
0.64 kW
Explanation:
The potential energy of a mass (M) at some height (h) is computed from ...
PE = Mgh
At 1 kg/liter, the available power is the rate at which that energy is available ...
(490 kg/min)×(1 min/(60 s))×(9.8 m/s²)(8 m) ≈ 640.3 kg·m²/s³
= 640.3 W
In kilowatts, that is 0.64 kW.
Answer:
43%
Explanation:
As the problem says that the antimony has only two isotopes, lets call each isotope the following:
x=abundance of isotope 121Sb
1-x=abundance of isotope 123Sb
And
Atomic weight of antimony = (isotopic mass of 121Sb*x)+(isotopic mass of 123Sb*(1-x))
Replacing values we have:
![121.757=(120.904x)+(122.902(1-x))](https://tex.z-dn.net/?f=121.757%3D%28120.904x%29%2B%28122.902%281-x%29%29)
Solving for x:
![121.757=120.904x+122.902-122.902x](https://tex.z-dn.net/?f=121.757%3D120.904x%2B122.902-122.902x)
![121.757=-1.998x+122.902](https://tex.z-dn.net/?f=121.757%3D-1.998x%2B122.902)
![121.757-122.902=-1.998x](https://tex.z-dn.net/?f=121.757-122.902%3D-1.998x)
![-1.145=-1.998x](https://tex.z-dn.net/?f=-1.145%3D-1.998x)
![x=\frac{-1.145}{-1.998}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-1.145%7D%7B-1.998%7D)
![x=0.57](https://tex.z-dn.net/?f=x%3D0.57)
it means that the abundance of the isotope 121Sb is 57% and the abundance of isotope 123Sb is 43%