Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15 Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15 Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample
<span>The liquid form of matter is usually more dense than its gas form. This is because liquid molecules are closer together compared to gas molecules. An exception, however, is water. Water's solid form or ice is less dense than its liquid form because of the orientation of hydrogen bonds that lowers its density.</span>
Atomic weight is actually calculated by the sum of protons and neutrons of that atom. It is not equal to the number. You must add them for the result
The atom has equal amount of Protons and electrons it is Neutral
Answer: The approximate equilibrium partial pressure of
is 3.92 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
The given balanced equilibrium reaction is,

![K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
![1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
On reversing the reaction:

initial pressure 4.00atm 2.00 atm 0
eqm (4.00-2x)atm (2.00-x) atm 2x atm
![K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)


![0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}](https://tex.z-dn.net/?f=0.67%5Ctimes%2010%5E5%3D%5Cfrac%7B2x%5D%5E2%7D%7B%5B4.00-2x%5D%5E2%5Ctimes%20%5B2.00-x%5D%7D)

![[H_2S]=2x=2\times 1.96=3.92 atm](https://tex.z-dn.net/?f=%5BH_2S%5D%3D2x%3D2%5Ctimes%201.96%3D3.92%20atm)
Thus approximate equilibrium partial pressure of
is 3.92 atm