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
To determine the number of moles of carbon dioxide that is produced, we need to know the reaction of the process. For the reaction of HCl and sodium carbonate, the balanced chemical equation would be expressed as:
2HCl + Na2CO3 = 2NaCl + H2O + CO2
From the initial amount given of sodium carbonate and the relation of the substances from the balanced reaction, we calculate the moles of carbon dioxide as follows:
0.2 moles Na2Co3 ( 1 mol CO2 / 1 mol Na2Co3 ) = 0.2 moles CO2
Therefore, the amount in moles of carbon dioxide that is produced from 0.2 moles sodium carbonate would be 0.2 moles as well.
Answer:
Explanation:The atomic number of sodium is 11. That is, the number of electrons in sodium is 11. Therefore, a sodium atom will have two electrons in the first shell, eight in the 2nd orbit, and an electron in the 3rd shell.
Answer:
41.9 g
Explanation:
We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.
Q = c × m × ΔT
where,
c: specific heat capacity
m: mass
ΔT: change in temperature
If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.
According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.
Qw + Qs = 0
Qw = -Qs
cw × mw × ΔTw = -cs × ms × ΔTs
(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)
ms = 41.9 g
