Answer: 1.414x10^24 molecules in 94.4g MgO
Explanation: molar mass MgO 40.204
molecules in 40.204 g MgO = avogadro number
molecules in 94.4 g MgO = (94.4/40.204)*avogadro number
(94.4/40.204)*6.02214076*10^23 = 14.14x10^23
We see that in the left-hand side of the equation, the side of the reactants, that we have 2 moles of Na and bromine is in it's diatomic form.
Therefore, we have 2 moles of each of these elements. When we combine the sodium bromide molecule in the products, we are going to want to keep these molar amounts the same. So we are going to need to put a 2 in front of the sodium bromide in order to correctly balance this equation.
So the coefficient for sodium bromide (NaBr) in this equation is 2.
<span>1.190x10^-1 m/l
First, calculate how many moles of NaOH was used.
0.03430 l * 8.670x10^-2 mol/l = 2.97381x10^-3 mol
So we used 2.97381x10^-3 moles of NaOH, which means we had 2.97381x10^-3 moles of HCl in the original sample. So the molarity of the original sample is the number of moles divided by the volume.
2.97381x10^-3 mol / 0.02500 = 0.1189524 m
Rounding to 4 significant figures gives 0.1190 or 1.190x10^-1 m</span>