The question is incomplete, here is the complete question:
Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine.
at 1,000 K
Calculate Kc for the reaction ![2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)](https://tex.z-dn.net/?f=2CCl_4%28g%29%2BO_2%28g%29%5Crightleftharpoons%202COCl_2%28g%29%2B2Cl_2%28g%29)
<u>Answer:</u> The value of
for the final reaction is ![1.936\times 10^{19}](https://tex.z-dn.net/?f=1.936%5Ctimes%2010%5E%7B19%7D)
<u>Explanation:</u>
The given chemical equations follows:
![CCl_4(g)+\frac{1}{2}O_2(g)\rightleftharpoons COCl_2(g)+Cl_2(g);K_c](https://tex.z-dn.net/?f=CCl_4%28g%29%2B%5Cfrac%7B1%7D%7B2%7DO_2%28g%29%5Crightleftharpoons%20COCl_2%28g%29%2BCl_2%28g%29%3BK_c)
We need to calculate the equilibrium constant for the equation, which is:
![2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)](https://tex.z-dn.net/?f=2CCl_4%28g%29%2BO_2%28g%29%5Crightleftharpoons%202COCl_2%28g%29%2B2Cl_2%28g%29)
As, the final reaction is the twice of the initial equation. So, the equilibrium constant for the final reaction will be the square of the initial equilibrium constant.
The value of equilibrium constant for net reaction is:
![K_c'=(K_c)^2](https://tex.z-dn.net/?f=K_c%27%3D%28K_c%29%5E2)
We are given:
![K_c=4.4\times 10^9](https://tex.z-dn.net/?f=K_c%3D4.4%5Ctimes%2010%5E9)
Putting values in above equation, we get:
![K_c'=(4.4\times 10^9)^2=1.936\times 10^{19}](https://tex.z-dn.net/?f=K_c%27%3D%284.4%5Ctimes%2010%5E9%29%5E2%3D1.936%5Ctimes%2010%5E%7B19%7D)
Hence, the value of
for the final reaction is ![1.936\times 10^{19}](https://tex.z-dn.net/?f=1.936%5Ctimes%2010%5E%7B19%7D)
Answer:
40 Grams
Explanation:
Fe2O3 is 2 parts Iron 3 parts Oxygen.
Assuming the molecular weights of both are equivalent (<em>ideal elemental law</em>), it is a 2:3 ratio, or 2/5 of the total.
2/5 of 100 is 40, so the answer is
40 grams
Answer:
![\boxed{\text{115.2 torr}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B115.2%20torr%7D%7D)
Explanation:
Let’s call water Component 1 and lactose Component 2.
According to Raoult’s Law,
![p_{1} = \chi_{1}p_{1}^{\circ} \text{ and}\\p_{2} = \chi_{2}p_{2}^{\circ}](https://tex.z-dn.net/?f=p_%7B1%7D%20%3D%20%5Cchi_%7B1%7Dp_%7B1%7D%5E%7B%5Ccirc%7D%20%5Ctext%7B%20and%7D%5C%5Cp_%7B2%7D%20%3D%20%5Cchi_%7B2%7Dp_%7B2%7D%5E%7B%5Ccirc%7D)
where
p₁ and p₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole fractions of the components
p₁° and p₂° are the vapour pressures of the pure components.
Data:
m₁ = 110.0 g; p₁° = 118.0 torr
m₂ = 50.00 g; p₂° = 0 torr
1. Calculate the moles of each component
![n_{1} = \text{110.0 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{6.104 mol}\\\\n_{2} = \text{50.00 g} \times \dfrac{\text{1 mol}}{\text{342.3 g}} = \text{0.1461 mol}](https://tex.z-dn.net/?f=n_%7B1%7D%20%3D%20%5Ctext%7B110.0%20g%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%7D%7D%7B%5Ctext%7B18.02%20g%7D%7D%20%3D%20%5Ctext%7B6.104%20mol%7D%5C%5C%5C%5Cn_%7B2%7D%20%3D%20%5Ctext%7B50.00%20g%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%7D%7D%7B%5Ctext%7B342.3%20g%7D%7D%20%3D%20%5Ctext%7B0.1461%20mol%7D)
2. Calculate the mole fraction of each component
![\begin{array}{rcl}\chi_{2} & = & \dfrac{n_{2}}{n_{1} + n_{2}}\\\\&= & \dfrac{0.1461}{6.104 + 0.1461}\\\\& = &\dfrac{0.1461}{6.250}\\\\& = & \mathbf{0.023 37}\\\chi_{1}& = &1 - \chi_{2}\\& = &1 - 0.023 37\\& = & \mathbf{0.976 63}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5Cchi_%7B2%7D%20%26%20%3D%20%26%20%5Cdfrac%7Bn_%7B2%7D%7D%7Bn_%7B1%7D%20%2B%20n_%7B2%7D%7D%5C%5C%5C%5C%26%3D%20%26%20%5Cdfrac%7B0.1461%7D%7B6.104%20%2B%200.1461%7D%5C%5C%5C%5C%26%20%3D%20%26%5Cdfrac%7B0.1461%7D%7B6.250%7D%5C%5C%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B0.023%2037%7D%5C%5C%5Cchi_%7B1%7D%26%20%3D%20%261%20-%20%5Cchi_%7B2%7D%5C%5C%26%20%3D%20%261%20-%200.023%2037%5C%5C%26%20%3D%20%26%20%5Cmathbf%7B0.976%2063%7D%5C%5C%5Cend%7Barray%7D)
3. Calculate the vapour pressure of the mixture
![p_{1} = 0.97663 \times \text{118.0 torr} = \text{ 115.2 torr}\\p_{2} = 0\\p_{\text{tot}} = p_{1} + p_{2} = \text{115.2 torr + 0} = \textbf{115.2 torr}\\\text{The partial pressure of water above the solution is $\boxed{\textbf{115.2 torr}}$}](https://tex.z-dn.net/?f=p_%7B1%7D%20%3D%200.97663%20%5Ctimes%20%5Ctext%7B118.0%20torr%7D%20%3D%20%5Ctext%7B%20115.2%20torr%7D%5C%5Cp_%7B2%7D%20%3D%200%5C%5Cp_%7B%5Ctext%7Btot%7D%7D%20%3D%20p_%7B1%7D%20%2B%20p_%7B2%7D%20%3D%20%5Ctext%7B115.2%20torr%20%2B%200%7D%20%3D%20%5Ctextbf%7B115.2%20torr%7D%5C%5C%5Ctext%7BThe%20partial%20pressure%20of%20water%20above%20the%20solution%20is%20%24%5Cboxed%7B%5Ctextbf%7B115.2%20torr%7D%7D%24%7D)
The thing that they all have in common is that they all use electrical energy
Answer:
7.50×10¯² lbs = 7.50×10¯² lbs × 453.952 g / 1 lbs
7.50×10¯² lbs = 34.0464 g.
Explanation:
From the question given, the following data were obtained:
1 lbs = 453.952 g
7.50×10¯² lbs =.?
Thus, we can obtain convert 7.50×10¯² lbs to grams as follow:
1 lbs = 453.952 g
Therefore,
7.50×10¯² lbs = 7.50×10¯² lbs × 453.952 g / 1 lbs
7.50×10¯² lbs = 34.0464 g
Therefore, 7.50×10¯² lbs is equivalent to 34.0464 g.