I think that it is A) Conduction
Answer:
552 g of LiNO₃
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mass of LiNO₃ =?
Next, we shall determine the number of mole of LiNO₃ in the solution. This can be obtained as follow:
Volume of solution = 2 L
Molarity of LiNO₃ = 4 M
Mole of LiNO₃ =?
Molarity = mole /Volume
4 = mole of LiNO₃ / 2
Cross multiply
Mole of LiNO₃ = 4 × 2
Mole of LiNO₃ = 8 moles
Finally, we shall determine the mass of of LiNO₃ needed to prepare the solution. This is can be obtained as follow:
Mole of LiNO₃ = 8 moles
Molar mass of LiNO₃ = 7 + 14 + (16×3)
= 7 + 14 + 48
= 69 g/mol
Mass of LiNO₃ =?
Mole = mass /Molar mass
8 = Molar mass of LiNO₃ /69
Cross multiply
Molar mass of LiNO₃ = 8 × 69
Molar mass of LiNO₃ = 552 g
Thus, 552 g of LiNO₃ is needed to prepare the solution.
Answer:
2 K3(PO4) aq + 3NiBr2 aq -------- 6 KBr aq + Ni3(PO4)2 s
Explanation:
