Formula for Barium Nitrate = Ba(NO3)2
Thus based on stoichiometry:
1 mole of Ba(NO3)2 contains 2 moles of NO3-
Therefore, concentration of nitrate ion NO3- would be = 2*0.240 = 0.480 M
Use the relation:
V1M1 = V2M2
V1 = V2M2/M1 = 0.500 L * 0.0800/0.480 = 0.0833 L
Thus, 0.0833 L or 83.3 ml solution of Ba(NO3)2 would be required.
Answer:
Melting
Explanation:
Once heat is added it will turn to a liquid
a . knowledge and existing theories .
b. falsifiable scope
750
Curious clouds.... ...