Answer:
![P_T=112.4torr](https://tex.z-dn.net/?f=P_T%3D112.4torr)
Explanation:
Hello there!
In this case, since these problems about gas mixtures are based off Dalton's law in terms of mole fraction, partial pressure and total pressure, we can write the following for hydrogen, we are given its partial pressure:
![P_{H_2}=x_{H_2}*P_T](https://tex.z-dn.net/?f=P_%7BH_2%7D%3Dx_%7BH_2%7D%2AP_T)
And can be solved for the total pressure as follows:
![P_T=\frac{P_{H_2}}{x_{H_2}}](https://tex.z-dn.net/?f=P_T%3D%5Cfrac%7BP_%7BH_2%7D%7D%7Bx_%7BH_2%7D%7D)
However, we first calculate the mole fraction of hydrogen by subtracting that of nitrogen to 1 due to:
![x_{H_2}+x_{N_2}=1\\\\x_{H_2}=1-0.333=0.667](https://tex.z-dn.net/?f=x_%7BH_2%7D%2Bx_%7BN_2%7D%3D1%5C%5C%5C%5Cx_%7BH_2%7D%3D1-0.333%3D0.667)
Then, we can plug in to obtain the total pressure:
![P_T=\frac{75.0torr}{0.667}\\\\P_T=112.4torr](https://tex.z-dn.net/?f=P_T%3D%5Cfrac%7B75.0torr%7D%7B0.667%7D%5C%5C%5C%5CP_T%3D112.4torr)
Regards!
I think the answer would be d but not 100% sure
Answer:
(a) x₁= 0.004 444; (b) y₁ = -0.9545; (c) x₂ = 0.001 905; (d) y₂ = -0.4541;
(e) rise = 0.5004; (f) run = -0.002 539; (g) slope = -197.1; (h) Eₐ = -1.64 kJ·mol⁻¹
Explanation:
This is an example of the Arrhenius equation:
![k = Ae^{-E_{a}/RT}\\\text{Take the ln of each side}\\\ln k = \ln A - \dfrac{E_{a}}{RT}\\\\\text{We can rearrange this to give}\\\ln k = -\left ( \dfrac{E_{a}}{R} \right )\dfrac{1}{T} + \ln A](https://tex.z-dn.net/?f=k%20%3D%20Ae%5E%7B-E_%7Ba%7D%2FRT%7D%5C%5C%5Ctext%7BTake%20the%20ln%20of%20each%20side%7D%5C%5C%5Cln%20k%20%3D%20%5Cln%20A%20-%20%5Cdfrac%7BE_%7Ba%7D%7D%7BRT%7D%5C%5C%5C%5C%5Ctext%7BWe%20can%20rearrange%20this%20to%20give%7D%5C%5C%5Cln%20k%20%3D%20-%5Cleft%20%28%20%5Cdfrac%7BE_%7Ba%7D%7D%7BR%7D%20%5Cright%20%29%5Cdfrac%7B1%7D%7BT%7D%20%2B%20%5Cln%20A)
Thus, if you plot ln k vs 1/T, you should get a straight line with slope = -Eₐ/R and a y-intercept = lnA
(a) x₁
x₁= 1/T₁ = 1/225 = 0.004 444
(b) y₁
y₁ = ln(k₁) = ln0.385 = -0.9545
(c) x₂
x₂= 1/T₂ = 1/525 = 0.001 905
(d) y₂
y₂ = ln(k₂) = ln0.635 = -0.4541
(e) Rise
Δy = y₂ - y₁ = -0.4541 - (-0.9545) = -0.4541 + 0.9545 = 0.5004
(f) Run
Δx = x₂ - x₁ = 0.001 905 - 0.004 444 = -0.002 539
(g) Slope
Δy/Δx = 0.5004/(-0.002 539) K⁻¹ = -197.1
(h) Activation energy
Slope = -Eₐ/R
Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol
Answer:
the packing efficiency is 52.36%
Explanation:
Given the data in the question;
simple cubic unit cell that contains one atom with a metallic radius of 175 pm;
we know that;
Edge length of Simple cubic (a) is related to radius of atom (r) as follows;
a = 2r
since radius r = 175 pm
we substitute
a = 2 × 175 pm
a = 350 pm
Now we get the volume unit;
Volume of unit cell = a³ = ( 350 pm ) = 42875000 pm³
Next we get Volume of sphere;
Volume of Sphere =
πr³
Volume occupied by 1 atom =
× π × ( 175 pm )³
=
× π × 5359375 pm³
= 22449297.5 pm³
Now, the packing efficiency = ( Volume occupied by 1 atom / Volume of unit cell ) × 100
we substitute;
packing efficiency = ( 22449297.5 pm³ / 42875000 pm³ ) × 100
= 0.523598 × 100
= 52.36%
Therefore, the packing efficiency is 52.36%