The domain of Eukarya contains eukaryotes, which are cells with nuclei.
Answer:
6.2846
Explanation:
Given that:-
Concentrations at equilibrium :-
![[CH_4]=0.126\ M](https://tex.z-dn.net/?f=%5BCH_4%5D%3D0.126%5C%20M)
![[H_2O]= 0.242\ M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%200.242%5C%20M)
![[CO]= 0.126\ M](https://tex.z-dn.net/?f=%5BCO%5D%3D%200.126%5C%20M)
![[H_2]= 1.15\ M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%201.15%5C%20M)
The equilibrium reaction is:-
The expression for equilibrium constant is:
![K_{c}=\frac {\left [ H_2 \right ]^3\left [ CO \right ]}{\left [ CH_4 \right ]\left [ H_2O \right ]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H_2%20%5Cright%20%5D%5E3%5Cleft%20%5B%20CO%20%5Cright%20%5D%7D%7B%5Cleft%20%5B%20CH_4%20%5Cright%20%5D%5Cleft%20%5B%20H_2O%20%5Cright%20%5D%7D)
Applying the values as:-
<u>The equilibrium constant for the reaction is:- 6.2846</u>
Answer:
C11H25SO4
Explanation:
The total mass of the compound is 253.4 g, so, the mass of each element will be:
C: 52.14% of 253.4 = 0.5214x253.4 = 132.12 g
H: 9.946% of 253.4 = 0.09946x253.4 = 25.20 g
S: 12.66% of 253.4 = 0.1266x253.4 = 32.08 g
O: 25.26% of 253.4 = 0.2526x253.4 = 64.00 g
The molar mass are: C = 12 g/mol, H 1 g/mol, S = 32 g/mol, and O = 16 g/mol
So, to know how much moles will be, just divide the mass calculated above for the molar mass:
C: 132.12/12 = 11 moles
H: 25.20/ 1 = 25 moles
S: 32.08/32 = 1 mol
O: 64.00/16 = 4 moles
So the molecular formula is C11H25SO4
C
This is because 10+5=15
15/45=0.3
