Answer:
B. 4 ml
Explanation:
Volume displacement of the hammer = volume volume of water when the hammer was placed into the cylinder - volume of water only before the hammer was placed into the cylinder.
Volume volume of water when the hammer was placed into the cylinder = 69 ml
Volume of water only before the hammer was placed into the cylinder = 65 ml
Volume displacement of the hammer = 69 ml - 65 ml = 4 ml
You have to use Avogadro's number (6.02x10^23 molecules/mole) to find the number of moles each reactant starts off with.
moles of Fe and O₂:
12 atoms/(6.02x10^23 atoms/mole)=1.99x10^-23 mol Fe
6 molecules/(6.02x10^23 molecules/mole)=9.967x10^-24 mol <span>O₂
</span>Then you find the limiting reagent by finding how much product each given amount of reactant can make. Which ever one produces the least amount of product is the limiting reagent.
amount of Fe₂O₃ produced:
<span>(1.99x10^-23 mol Fe)x(2mol/4mol)= 9.967x10^-24mol Fe</span>₂O₃<span>
</span>(9.967x10^-24 mol O₂)x(2mol/3mol)= 6.645x10^-24 mol Fe₂O₃<span>
</span>since oxygen produces the leas amount of product, oxygen is the limiting reagent. since we know that oxygen is the limiting reagent we can use the amount of product formed with oxygen to find the amount of iron used.
6.645x10^-24 mol Fe₂O₃x(4mol/2mol)=1.329x10^-23 mol Fe consumed
<span> find the amount left over by subtracting the original amount of Fe by the amount consumed in the reaction.
</span>1.993x10^-23-1.329x10^-23= 6.645x10^-23mol Fe left
find the number of atoms by multiplying that by Avogadro's number.
<span>(6.645x10^-23mol)x(6.02x10^23 atoms/mol)=4 atoms
</span>therefore 4 atoms of Fe will be left over after the reaction happens.
I hope this helps.
step one
calculate the % of oxygen
from avogadro constant
1moles = 6.02 x 10 ^23 atoms
what about 4.33 x10^22 atoms
= ( 4.33 x 10^ 22 x 1 mole ) / 6.02 10^23= 0.0719 moles
mass= 0.0719 x16= 1.1504 g
% composition is therefore= ( 1.1504/3.25) x100 = 35.40%
step two
calculate the % composition of chrorine
100- (25.42 + 35.40)=39.18%
step 3
calculate the moles of each element
that is
Na = 25.42 /23=1.1052 moles
Cl= 39.18 /35.5=1.1037moles
O= 35.40/16= 2.2125 moles
step 4
find the mole ratio by dividing each mole by 1.1037 moles
that is
Na = 1.1052/1.1037=1.001
Cl= 1.1037/1.1037= 1
0=2.2125 = 2
therefore the empirical formula= NaClO2
Answer:
7
Explanation:
The mass number is neutrons plus protons
Answer:
A). 92.02g
Explanation:
Equation of the reaction;
N2 (g)+ 2O2(g)------> 2NO2(g)
Note that the balanced reaction equation is the first step in solving any problem on stoichiometry. Once the reaction equation is correct, the question can be easily solved.
Reaction of one mole of nitrogen gas with two moles of oxygen gas yields two moles of nitrogen dioxide.
Mass of two moles of nitrogen dioxide= 2[14 + 2(16)] = 2[14+32]= 2[46]= 92 gmol-1
Therefore; Mass of two moles of nitrogen dioxide is 92
