It’s [Xe]7s^2 5f^14 (first one) hopefully i could help!!
Explanation:
The given reaction will be as follows.
So, equilibrium constant for this equation will be as follows.
![K_{c} = \frac{[CH_{3}OH]}{[CO][H_{2}]^{2}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCH_%7B3%7DOH%5D%7D%7B%5BCO%5D%5BH_%7B2%7D%5D%5E%7B2%7D%7D)
As it is given that concentration of all the species is 2.4. Therefore, calculate the value of equilibrium constant as follows.
= 
= 0.173
Thus, we can conclude that equilibrium constant for the given reaction is 0.173.
This would be measurements.
Answer:
Zn3P2O8
Explanation:
In this particular question, it is necessary to convert the respective masses to percentages. We convert to percentages by placing each mass over the total mass and multiplying by 100%. Since the total is 50mg, conversion to percentage can be done by multiplying the masses by 2 as 100/50 is 2
For Oxygen = 16.58 * 2 = 33.16%
For phosphorus = 8.02 * 2 = 16.04%
For zinc = 25.40 * 2 = 50.80%
We then proceed to divide these percentages by their respective atomic masses. The atomic mass of oxygen, phosphorus and zinc are 16, 31 and 65 respectively.
O = 33.16/16 = 2.0725
P = 16.04/31 = 0.5174
Zn = 50.80/65 = 0.7815
Now, we divide by the smallest value which is that of the phosphorus
O = 2.0725/0.5174 = 4
P = 0.5174/0.5174 = 1
Zn= 0.7815/0.5174 = 1.5
Now, we need to multiply through by 2. This yields: O = 8, P = 2 and Zn = 3
The empirical formula is thus: Zn3P2O8
