Answer:
[Ba^2+] = 0.160 M
Explanation:
First, let's calculate the moles of each reactant with the following expression:
n = M * V
moles of K2CO3 = 0.02 x 0.200 = 0.004 moles
moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles
Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.
Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3
As you can see, 0.04 moles of K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of
0.012 - 0.004 = 0.008 moles of Ba(NO3)2
These moles are in total volume of 50 mL (30 + 20 = 50)
So finally, the concentration of Ba in solution will be:
[Ba] = 0.008 / 0.050 = 0.160 M
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent Bond
Between 0.4 and 1.7 then it is Polar Covalent Bond
Greater than 1.7 then it is Ionic
For N and Se,
E.N of Nitrogen = 3.04
E.N of Selenium = 2.55
________
E.N Difference
0.49 (Weakly Polar Covalent)
As you have not provided remaining pairs, so if you have any of them, follow the method as mentioned above.
The number of collison is independent of volume
