Answer:
Partial pressure of of CO₂ in the product mixture is 0,20atm
Explanation:
The balance equation is:
2CO(g) + O₂(g) → 2CO₂(g)
Total pressure of CO(g) and O₂(g) gases before reaction at 100,0°C and 1,0L is 1,50 atm. You can say:
X₁ + Y₁ = 1,50atm <em>(1)</em>
Where X₁ is initial partial pressure CO and Y₁ is initial partial pressure of O₂
After reaction partial pressures are:
X₂ = X₁ - 2n = 0; <em>2n = X₁</em>
Y₂ = Y₁ - n
Z₂ = 2n
Where Z₂ is final partial pressure of CO₂
After reaction pressure at 100,0°C and 1,0L is 1,40 atm, that means:
1,40 atm = (Y₂ + Z₂)
1,40 atm = Y₁ - n + 2n
1,40atm = Y₁ + n
1,40 atm = Y₁ + X₁/2 <em>(2)</em>
Replacing (1) in (2)
1,40 atm = 1,50atm - X₁ + X₁/2
-0,10 atm = - X₁/2
<em>0,20 atm = X₁</em>.
As 2n = X₁; 2n =<em> Z₂ = 0,20 atm</em>
I hope it helps!
It is going to be B the objects mass
Explanation:
The given data is as follows.
= 100 mm Hg or 
= 0.13157 atm
= 
= (1080 + 273) K = 1357 K
= 
= (1220 + 273) K = 1493 K
= 600 mm Hg or 
= 0.7895 atm
R = 8.314 J/K mol
According to Clasius-Clapeyron equation,
![log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]](https://tex.z-dn.net/?f=log%28%5Cfrac%7B0.7895%7D%7B0.13157%7D%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B2.303%20%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B1357%20K%7D%20-%20%5Cfrac%7B1%7D%7B1493%20K%7D%5D)
![log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]](https://tex.z-dn.net/?f=log%20%286%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B19.147%7D%5B%5Cfrac%7B%281493%20-%201357%29%20K%7D%7B1493%20K%20%5Ctimes%201357%20K%7D%5D)
0.77815 = 
= 
J/mol
= 
= 221.9 kJ/mol
Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.
Find one mole
8 C = 8 * 12 = 96
9 H = 1 * 9 = 9
4 O = 4 *16 = 64
Total = 169
1 mol = 169 grams.
0.432 mol = x
1/0.432 = 169/x
x = 0.432 * 169
x = 73.0 grams
Answer:
2,3–dimethylpentane
Explanation:
To know which option is correct, we shall determine the name of the compound.
To obtain the name of the compound, do the following:
1. Determine the longest continuous carbon chain. This gives the parent name of the compound.
2. Identify the substituent group attached to the compound.
3. Locate the position of the substituent group by giving it the lowest possible count.
4. Combine the above to obtain the name of the compound.
Now, we shall determine the name of the compound as follow:
1. The longest continuous carbon chain is 5. Thus, the parent name of the compound is pentane.
2. The substituent group attached is methyl (–CH₃)
3. There are two methyl group attached to the compound. One is located at carbon 2 and the other at carbon 3.
4. Therefore, the name of the compound is:
2,3–dimethylpentane
None of the options are correct.
