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)
Answer:
264 g/mol
Explanation:
#electrons equal #protons = 106
Plus 1 charge => m protons = 106 + 1 = 107
Mass number: 107 + 157 = 264 g/mol
The relation between rate constants at different temperatures, temperature and activation energy is known as Arrhenius equation
<u>Arrhenius equation</u>
For two temperatures
Where
Ea = ? = activation energy
k1 = 3.36 × 10⁴
T1=344 k
k2=7.69
T2=219K
R= gas constant = 8.314 J /molK
Putting values
Ea = (-69.69)/(-0.00166) = 41981.93 J/mol
Or
Activation energy is 42.0 kJ /mol
