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
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
Molar mass of water 18g/mol
Number of mols = 50.0g/18g/mol =2.78 mol
Heat absorbed = 40.7 kj/mol * 2.78 mol = 113.1 kj.
Answer:
Molarity is halved when the volume of solvent is doubled.
Explanation:
Using the dilution equation (volume 1)(molarity 1)=(volume 2)(molarity 2), we can demonstrate the effects of doubling volume.
Suppose the starting volume is 1 L and the starting molarity is 1 M, and doubling the volume would make the final volume 2 L.
Plugging these numbers into the equation, we can figure out the final molarity.
(1 L)(1 M)=(2 L)(X M)
X M= (1 L x 1 M)/(2 L)
X M= 1/2 M
This shows that the molarity is halved when the volume of solvent is doubled.
Answer:
2.1 x 10^-2
Explanation:
Divide 8.4 x 10^-3 x 8.4 x 10^-3 by 5.82 x 10^-2 and you end up with 2.1 x 10^-2
