Answer:
Final temperature = 1279.25 K
Explanation:
We can solve this using the formula for Charles law since we are given volume and temperature.
From Charles law, we know that;
V1/T1 = V2/T2
Where;
T1 is the initial temperature
V1 is the initial volume
T2 is the final temperature
V2 is the final volume
We are given;
V1 = 2 L
T1 = 301 K
V2 = 8.5 L
Thus, making T2 the subject, we have;
T2 = V2•T1/V1
Plugging in the relevant values;
T2 = 8.5 × 301/2
T2 = 1279.25 K
Answer:
Average atomic mass = 15.86 amu.
Explanation:
Given data:
Number of atoms of Z-16.000 amu = 205
Number of atoms of Z-14.000 amu = 15
Average atomic mass = ?
Solution:
Total number of atoms = 205 + 15 = 220
Percentage of Z-16.000 = 205/220 ×100 = 93.18%
Percentage of Z-14.000 = 15/220 ×100 = 6.82 %
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (93.18×16.000)+(6.82×14.000) /100
Average atomic mass = 1490.88 + 95.48 / 100
Average atomic mass = 1586.36 / 100
Average atomic mass = 15.86 amu.
Answer:
Explanation:
We know, 
where, R = 0.0821 L.atm/(mol.K), T is temperature in kelvin and 
is difference in sum of stoichiometric coefficient of products and reactants
Here 
and T = 311 K
So, ![K_{p}=(0.0111)\times [(0.0821L.atm.mol^{-1}.K^{-1})\times 311K]^{-1}=4.35\times 10^{-4}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D%280.0111%29%5Ctimes%20%5B%280.0821L.atm.mol%5E%7B-1%7D.K%5E%7B-1%7D%29%5Ctimes%20311K%5D%5E%7B-1%7D%3D4.35%5Ctimes%2010%5E%7B-4%7D)
Hence value of equilibrium constant in terms of partial pressure 
is 
Answer:
74,67 gr/mol
Explanation:
At STP 1 mole of an ideal gas has volume of 22,4 L. Since we know the volume of the gas we can find the number of moles of the gas. (300 mL=0,3 L)
n=0,3L/22,4 L=0,01339 mol
Since we know weight of the gas as 1 g, we can find the molecular weight as;
MW=1 g/0,01339 mol =74,67 gr/mol
