Answer:
0.32M
Explanation:
<u>Step 1:</u> Balance the reaction
K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3
We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution
SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3
<u>Step 2: </u>Calculate concentration
To find the concentration of the barium cation we use the following equation:
Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>
<u />
<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3
[Ba2+] = 0.32 M
The concentration of Barium ion in solution is 0.32 M
answer an element and a compound
Answer:
was there a question you needed help with?
I think the correct answer is C
P1 * V1 ÷ T1 = P2 * V2 ÷ T2
45 * 1.20 ÷ 314 = 96 * V2 ÷ 420
30,144 * V2 = 22,680
V2 = 22,680 ÷ 30,144
The new volume is approximately 0.75 liter.
I hope I helped
