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.
Losing eltron is the answer
34g C * ( 1 mol / 12.0107 ) * ( 1 mol H2 / 1 mol C ) * ( <span>2.01588 g / 1 mol H2 ) = 5.70657164028741 g H2 = 5.7 g H2
Convert grams of C to moles of C using the given amount of grams and the molar mass ( 12.0107 g/mol ).
Gather the mole ratio from the coefficients in the balanced equation and multiply by the ratio.
Convert moles of H2 to grams of H2 </span> using the given amount of grams and the molar mass ( 2.01588 g/mol )<span>.
Revise your answer to have the correct number of significant figures. </span>
