3Mg + N₂= Mg₃N₂
n(Mg)=12,2g÷<span>24,4g/mol=0,5mol-limiting reagent
</span>n(N₂)=5,16g÷28g/mol=0,18mol
n(Mg₃N₂):n(Mg)=1:3, n(Mg₃N₂)=0,166mol, m(Mg₃N₂)=0,166·101,2=16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2=27,66%=0,2766
%(Mg) = 3·Ar(Mg)÷Mr(Mg₃N₂)= 3·24,4÷101,2=72,34% or 100% - 27,66%= 72,34%.
Explanation:
9/4 Be +2 (the 9 and 4 are stacked next to Be). Atomic #: 4
Mass #: 9, # protons: 4, # neutrons: 5, #electrons: 2.
31/15 P (31 is stacked over 15 next to the P). Atomic #: 15,
Mass #: 31, # protons: 15, # neutrons: 16, # electrons: 15.
Sugar. (We need a design tech section)
Answer:
Formula: Na2S2O3
we get solubility.
Divide the mass of the compound by the mass of the solvent and then multiply by 100 g to calculate the solubility in g/100g .
Solution given:
mass of sodium thiosulphate [m1]=25.5g
mass of water [m2]=40g
at temperature [t]=25°C
we have
<u>solubility in g/dm^3</u> :
- =

- =63.75g /litre=63.75g/dm³
<u>solubility in g/dm^3 :63.75g/dm³</u>
<u>n</u><u>o</u><u>w</u>
solubility of the solute in mol/dm^3=:63.75g/dm³/178=0.4 mol/dm³