Answer:
I know you have been waiting awhile for this question to be answered :)
Stoichiometry is used in industry quite often to determine the amount of materials required to produce the desired amount of products in a given useful equation. Each one of these products requires stoichiometry. There would be no products from these industries without chemical stoichiometry.
Explanation:
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Sorry you had to wait so long :(
Answer:
Explanation:
Hello,
In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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Because if the the technique is wrong the scientist is wrong I’m sorry it’s a bad answer :(
Answer:
The correct answer is Pu, 234.
Explanation:
In the given case, let us consider the reactant as X. Now the mass number (balanced) on both the sides will be,
Mass of X = Mass of Molybdenum + Mass of Tin + Mass of neutrons
M = 1 * 103 + 1 * 131 + 2 * 0
M = 234
Now the atomic number (balanced) on both the sides,
Atomic number of X = Atomic number of Molybdenum + Atomic number of Tin + Atomic number of neutrons
A = 1*42 + 1*50 + 2*1
A = 94
The atomic number 94 is for the element Plutonium, whose symbol is Pu. Thus, the reactant is 234-Pu.
Molarity=Moles of solute/Volume of solution in L
So
- 0.56M=moles/2.5L
- moles=0.56(2.5)
- moles of Iodine=1.4mol
Mads of Iodine
- Moles(Molar mass)
- 1.4(126.9)
- 177.66g
