Their prey would decrease
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.
The mass of a textbook was really heavy.
Hope I helped
Answer:
1. 2+ (
).
2. 0 (
).
Explanation:
Hello,
In this case, the described chemical reaction is a redox reaction in fact, since the oxidation states of both magnesium and copper change as shown due to the displacement:
Therefore:
1. Since copper is the cation in the copper (II) nitrate, the (II) means that its charge is 2+ ().
2. Since copper is alone, it means no electrons are being neither shared not given, its charge is 0 ().
Best regards.
Answer:
Explanation:
Hello,
In this case, since silver is initially hot as it cools down, the heat it loses is gained by the liquid, which can be thermodynamically represented by:
That in terms of the heat capacities, masses and temperature changes turns out:
Since no phase change is happening. Thus, solving for the heat capacity of the liquid we obtain:
Best regards.
