Answer:
Explanation:
The pressure of a gaseous mixture is equal to the sum of the partial pressures of the individual gases:
Σ
![= P_1+P_2+P_3+...+P_n](https://tex.z-dn.net/?f=%3D%20P_1%2BP_2%2BP_3%2B...%2BP_n)
The prompt is trying to confuse you, but it actually tells us the pressure of the mixture to be 1 atm, but this can be converted to torr. Furthermore, we are informed only three gases are in the mixture: diatomic nitrogen, diatomic oxygen, and carbon dioxide:
![P_g_a_s=1 \ atm = 760 \ torr= P_N_2+P_O_2+P_C_O_2\\760 \ torr = 582.008 \ torr + P_O_2 \ + 0.285 \ torr](https://tex.z-dn.net/?f=P_g_a_s%3D1%20%5C%20atm%20%3D%20760%20%5C%20torr%3D%20P_N_2%2BP_O_2%2BP_C_O_2%5C%5C760%20%5C%20torr%20%3D%20582.008%20%5C%20torr%20%2B%20P_O_2%20%5C%20%2B%200.285%20%5C%20torr)
Solve for Po2:
![P_o_2=(760-582.008-0.285) \ torr = 177.707 \ torr](https://tex.z-dn.net/?f=P_o_2%3D%28760-582.008-0.285%29%20%5C%20torr%20%3D%20177.707%20%5C%20torr)
Thus, the partial pressure of diatomic oxygen is 177.707 torr.
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Answer: Definitely the answer is C. They both last for the same number of days.
Explanation: This requires the moon to complete a bit more than one orbit to be between the earth and the sun again. The period from new moon to full moon is one half a synodic month of 29.530/2 which is 14.765 days.
<u>Answer:</u> The value of
for the reaction is
<u>Explanation:</u>
We are given:
Initial moles of ![NH_3=0.0150mol](https://tex.z-dn.net/?f=NH_3%3D0.0150mol)
Initial moles of ![O_2=0.0150mol](https://tex.z-dn.net/?f=O_2%3D0.0150mol)
Volume of the container = 1.00 L
Molarity of the solution = ![\frac{\text{Number of moles}}{\text{Volume of container}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BNumber%20of%20moles%7D%7D%7B%5Ctext%7BVolume%20of%20container%7D%7D)
![[NH_3]_i=\frac{0.0150}{1.00}=0.0150M](https://tex.z-dn.net/?f=%5BNH_3%5D_i%3D%5Cfrac%7B0.0150%7D%7B1.00%7D%3D0.0150M)
![[O_2]_i=\frac{0.0150}{1.00}=0.0150M](https://tex.z-dn.net/?f=%5BO_2%5D_i%3D%5Cfrac%7B0.0150%7D%7B1.00%7D%3D0.0150M)
The given chemical equation follows:
![4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)](https://tex.z-dn.net/?f=4NH_3%28g%29%2B3O_2%28g%29%5Crightarrow%202N_2%28g%29%2B6H_2O%28g%29)
Initial: 0.0150 0.0150
At eqllm: 0.0150-4x 0.0150-3x 2x 6x
The expression of
for above equation follows:
.......(1)
We are given:
Equilibrium concentration of ![N_2=1.96\times 10^{-3}](https://tex.z-dn.net/?f=N_2%3D1.96%5Ctimes%2010%5E%7B-3%7D)
Equating the equilibrium concentrations of nitrogen, we get:
![2x=1.96\times 10^{-3}\\\\x=0.98\times 10^{-3}M](https://tex.z-dn.net/?f=2x%3D1.96%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5Cx%3D0.98%5Ctimes%2010%5E%7B-3%7DM)
Calculating the equilibrium concentrations:
Concentration of ![NH_3=(0.0150-4x)=0.0150-4(0.00098)=0.01108M](https://tex.z-dn.net/?f=NH_3%3D%280.0150-4x%29%3D0.0150-4%280.00098%29%3D0.01108M)
Concentration of ![O_2=(0.0150-3x)=0.0150-3(0.00098)=0.01206M](https://tex.z-dn.net/?f=O_2%3D%280.0150-3x%29%3D0.0150-3%280.00098%29%3D0.01206M)
Concentration of ![N_2=2x=2(0.00098)=0.00196M](https://tex.z-dn.net/?f=N_2%3D2x%3D2%280.00098%29%3D0.00196M)
Concentration of ![H_2O=6x=6(0.00098)=0.00588M](https://tex.z-dn.net/?f=H_2O%3D6x%3D6%280.00098%29%3D0.00588M)
Putting values in expression 1, we get:
![K_c=\frac{(0.00196)^2\times (0.00588)^6}{(0.01108)^4\times (0.01206)^3}\\\\K_c=6.005\times 10^{-6}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%280.00196%29%5E2%5Ctimes%20%280.00588%29%5E6%7D%7B%280.01108%29%5E4%5Ctimes%20%280.01206%29%5E3%7D%5C%5C%5C%5CK_c%3D6.005%5Ctimes%2010%5E%7B-6%7D)
Hence, the value of
for the reaction is
Hi,
Answer is 191.2.
800J = 191.2 cal
Hope this helps.
Answer:
the answer is c hope this helps
Explanation: