830 mL. A 2.3 mol/L solution of CaCl2 has a volume of 830 mL
I am guessing that the concentration of your solution is 2.3 mol/L.
a) Moles of CaCl2
MM of CaCl2 = 110.98 g/mol
Moles of CaCl2 = 212 g CaCl2 x (1 mol CaCl2/110.98 g CaCl2)
= 1.910 mol CaCl2
b) Volume of solution
V = 1.910 mol CaCl2 x (1 L solution/2.3 mol CaCl2) = 0.83 L solution
= 830 mL solution
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.
Answer:
78% nitrogen
Explanation:
It has been long agreed upon that our atmosphere is highly abundant with nitrogen. Our atmosphere is composed of 78% nitrogen and then other gases follow such as 21% oxygen, 0.9% argon, and 0.03% carbon dioxide with very small percentages of other elements.
Answer:
answer is C
Explanation:
encourage the release of carbon dioxide from the stems
