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:
44
Explanation:
Given that :
Mass of solute = Mass of urea = 16g
Mass of water = 20g
Mass of solution = (mass of solute + mass of solvent) = (mass of urea + mass of water) = (16g + 20g) = 36g
Percentage Mass = (mass of solute / mass of solution) * 100%
Percentage Mass = (16 / 36) * 100%
Percentage Mass = 0.4444444 x 100%
Percentage Mass = 44.44%
Percentage Mass = 44%
Standard temperature is 273 K
Standard pressure is 1 atm
We use the ideal gas equation to find out density of nitrogen gas in g/L
Ideal gas equation:
Molar mass of 
Pressure = 1 atm
Temperature = 273 K
= 1.25 g/L
Therefore, density of nitrogen gas at STP is 1.25 g/L
<span>q(rxn) = - [q(water)+q(bomb)]
q(rxn) = -{[ (1000g)(4.184)(5.0)] + [ (5.0)(0.10)]}
q(rxn) = - (20920 + 0.5)
Now we divide 3.5g
q(rxn)= - (20920)/(3.5g)
q(rxn) = 5977.14
And final answer, change is to Kilo joule unit
-q(rxn) = 5.23 KJ/unit</span>
Answer:
b. The molarity of the solution increases
Explanation:
The correct answer is option b, that is the molarity of the solution increases.
Because the molarity is the concentration of the solution and it is explained as the amount of solute in amount of solution.
Solution: is the solute dissolved in solvent.
So if we increases the amount of solute in solvent the concentration in terms of molarity of solution increases and if we increase amount of solvent or water then the concentration or molarity increases.
Suppose we have form a sugar solution of 1 L by adding 4 mole of sugar then what happen
Use the Molarity formula
Molarity = no. of moles / 1 L of solution
put values in the formula
Molarity = 4/ 1 L of solution = 4 M
So the molarity of solution is 4 now if we add 2 mole more sugar to the same amount of sugar and amount of solution remain the same
now the no. of moles of sugar = 6 mole
So,
Use the Molarity formula
Molarity = no. of moles / 1 L of solution
put values in the formula
Molarity = 6 mol / 1 L of solution = 6 M
So the correct option is b.
