You can solve this problem through dimensional analysis.
First, find the molar mass of NaHCO3.
Na = 22.99 g
H = 1.008 g
C = 12.01 g
O (3) = 16 (3) g
Now, add them all together, you end with with the molar mass of NaHCO3.
22.99 + 1.008 + 12.01 + 16(3) = 84.008 g NaHCO3. This number means that for every mole of NaHCO3, there is 84.008 g NaHCO3. In simpler terms, 1 mole NaHCO3 = 84.008 g NaHCO3.
After finding the molar mass of sodium bicarbonate, now you can use dimensional analysis to solve for the number of moles present in 200. g of sodium bicarbonate.
Cross out the repeating units which are g NaHCO3, and the remaining unit is mole NaHCO3
200. * 1 = 200
200/ 84.008 = 2.38
Notice how there are only 3 sig figs in the answer. This is because the given problem only gave three sig figs.
Your final answer is 2.38 mol NaHCO3.
I think d is wrong because Everything else is correct for sure. also, minerals vary in the position they hold and have different chemical compounds.
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%.
Answer:
3
three half-filled orbitals each capable of forming a single covalent Bond and an additional lone - pair of electrons
The grams of NaCl that are required to make 150.0 ml of a 5.000 M solution is 43.875 g
calculation
Step 1:calculate the number of moles
moles = molarity x volume in L
volume = 150 ml / 1000 = 0.15 L
= 0.15 L x 5.000 M = 0.75 moles
Step 2: calculate mass
mass = moles x molar mass
molar mass of NaCl = 23 + 35.5 = 58.5 mol /L
mass is therefore =0.75 moles x 58.5 mol /l =43.875 g
