Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:

Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:


The concentration of
= x = 0.00212 M
Now we have to calculate the pH of solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


Therefore, the pH of the solution is, 2.67
1) Use the fact that 1 mol of gas at STP occupies 22.4 liter
=> 1 mol / 22.4 l = x / 0.125 l => x = 0.125 l * 1 mol / 22.4 l = 0.00558 mol
2) Now use the molar mass of the gas
molar mass of CO2 ≈ 44 g / mol
Formula: molar mass = mass in grams / number of moles =>
mass in grams = molar mass * number of moles = 44 g/mol * 0.00558 moles
mass = 0.246 g
Answer: 0.246 g
Answer:
5.004kg
Explanation:
Combustion of carbon
C+O2=CO2
from the relationship of molar ratio
mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)
mass of carbon =1000kg
atomic mass of carbon =12
volume of CO2 produced=1000×22.4/12
volume of CO2 produced =1866.6dm3
from the combustion reaction equation provided
CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)
applying the same relationship of molar ratio
no of mole of CO2=no of mole of urea
therefore
vol of CO2\22.4=mass of urea/molar mass of urea
molar mass of urea=60.06g/mol
from the first calculation
vol of CO2=1866.6dm3
mass of urea=1866.6×60.06/22.4
mass of urea=5004.82kg
Answer:
A
Explanation:
In all the other options either Carbon has more than 4 atoms attached or less than 4.
Carbon can not form more than four bonds. It can only share 4 electrons .
In other option C, No H-atom is linked in the right most Carbon.
whereas according to the definition of Hydrocarbon it should have been a Hydrogen atom.
Answer:
The answer would be C, in both chemical and physical change, because everything is made up of matter. So if you have a chemical or physical change, that doesent change the total mass of your product ever.
Explanation: