Molar mass magnesium bromide (MgBr2) = 184 g/mole
0.0939 mol/L x 0.720 L = 0.06761 moles MgBr2 required
0.06761 moles x 184 g/mole = 12.4 grams MgBr2 required (to 3 significant figures)
Answer:
A.) 8.796 B.) 234780 C.) 25.8
Explanation:
thats all i can read the rest is blurry
In gaseous particles are slightly lighter than those in pasma which look like liquid