Answer:
Look at my answer to the last one, and see if you can solve this one on your own!
After you try it, I can let you know if you got it right :)
Answer:
1. 266.22 g/mol
2. 168.81 g/mol
3. 223.35 g/mol
4. 199.88 g/mol
Explanation:
For you to calculate the molar mass of the salt you need to sum the molar masses of every element in the salt.
In the first salt, PdBr
, the subscript 2 means that there are 2 atoms of Br. So for you to calculate the molar mass of the salt you need to sum the molar mass of Pd and 2 times the molar mass of Br, as follows:
106 g/mol + 2(79.90 g/mol) = 266.22 g/mol
In the second salt BeBr there are 2 atoms of Br and 1 of Be, so the molar mass is:
9.012 g/mol +2(79.90 g/mol) = 186.22 g/mol
In the third salt CuBr there are 2 atoms of Br and 1 of Cu, so the molar mass is:
63.55 g/mol + 2(79.90 g/mol) = 223.35 g/mol
And in the fourth salt CaBr there are 2 atoms of Br and 1 of Ca, so the molar mass is:
40.08 g/mol + 2(79.90 g/mol) = 199.88 g/mol
Answer: 5.85kJ/Kmol.
Explanation:
The balanced equilibrium reaction is
The expression for equilibrium reaction will be,
![K_p=\frac{[p_{D}]\times [p_{C}]}^4{[p_{B}]^2\times [p_{A}]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5Bp_%7BD%7D%5D%5Ctimes%20%5Bp_%7BC%7D%5D%7D%5E4%7B%5Bp_%7BB%7D%5D%5E2%5Ctimes%20%5Bp_%7BA%7D%5D%7D)
Now put all the given values in this expression, we get the concentration of methane.
Relation of standard change in Gibbs free energy and equilibrium constant is given by:
where,
R = universal gas constant = 8.314 J/K/mole
T = temperature = 
= equilibrium constant = 10.6
Thus standard change in Gibbs free energy of this reaction is 5.85kJ/Kmol.
Answer:
33.3 kg of air
Explanation:
This is a problem of conversion unit.
Density is mass / volume
Therefore we have to calculate the volume in the room, to be multiply by density. That answer will be the mass of air.
Volume of the room → 9 ft . 11 ft . 10 ft = 990 ft³
Density is in g/L, therefore we have to convert the ft³ to dm³ (1 dm³ = 1L)
990 ft³ . 28.3 dm³ / 1ft³ = 28017 dm³ → 28017 L
This is the volume of the room, if we replace it in the density formula we can know the mass of air in g.
1.19 g/L = Mass of air / 28017 L
Mass of air = 28017 L . 1.19 g/L → 33340 g of air
Finally, let's convert the mass in g to kg → 33340 g . 1kg / 1000 g = 33.3 kg
