Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
PH + pOH = 14
11.8 + pOH = 14
pOH = 14 - 11.8
pOH = 2.2
[OH-] = 10 ^- pOH
[OH-] = 10 ^- 2.2
[OH-] = <span>6.33 x 10^-3 M
</span>
Answer B
hope this helps!
Well one characteristic gases and the state of matter(one of the distinct form i which matter exist)
I think 5.50 M x 35.0 mL x molar mass of RbOH = mass (g)
Answer:
Analytical Chemistry. The salt which in solution gives a pale green precipitate with sodium hydroxide solution and a white precipitate with barium chloride solution is : Iron (III) sulphate.
