2 1^/4 + 11^/2 = 9^/4 + 3^/2 = 9^/4 + 6^/4 = 15^/4 = 3 3^/4
If the products contain 3 nitrogen atoms, then so did the reactants since overall mass is conserved in a chemical reaction.
Answer: first blank is heating second blank is pressure i think but i know that its NOT freezing
Explanation:
I did it and got it right
830 mL. A 2.3 mol/L solution of CaCl2 has a volume of 830 mL
I am guessing that the concentration of your solution is 2.3 mol/L.
a) Moles of CaCl2
MM of CaCl2 = 110.98 g/mol
Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)
= 1.910 mol CaCl2
b) Volume of solution
V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution
= 830 mL solution
Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.
Answer: 43 L
