Answer:
NH3
Explanation:
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.
n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol
n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol
From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2
Can you give me the anwsers
Answer:
The answer is wind energy.
Explanation:
I did the quiz on time4learning.
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
First you have a knowledge of bond order which is
B.O=(no. of electrons in bonding orbital - no. of electrons in non-bonding orbital)÷2
Note:
bond strength is directly proportional to bond order.
For oxygen:
B.O=(6-2)/2= 2; after the removal of two electrons(removal occur from non-bonding orbital)
B.O=(6-0)/2= 3 (As B.O increased bond strength increased)
For Nitrogen:
B.O=(6-0)/2= 3; after the removal of two electrons(removal occur from bonding orbital)
B.O=(4-0)/2= 2 (As B.O decreased bond strength decreased)
