The first one is D, the second one is A and the last is C and D
35g Mg x 1mol / 24g = 840 mol
Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)
The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.
Wait is that suppose to be a question??!!
Molar mass of :
O2 = 16 * 2 = 32 g/mol
CO2 = 12 + 16 * 2 = 44 g/mol
<span>Balanced chemical equation :
</span>
1 CH4 +
2 O2 =
1 CO2 +
2 H2O
↓ ↓
2 moles 1 mole
2* 32 g O2 ----------> 1* 44 g CO2
x g O2 ------------> 10.0 g CO2
44 x = 2 * 32*10.0
44 x = 640


of O2