Why do you think it might be useful to know the distance between a gun fired and its target? (Forensics)
2 answers:
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For this question, we can use the molarity formula, which is as follows:
We know that we want a 0.1M solution in 100 mL, so we can plug those into the equation to find the number of moles required:
So now we know that the number of moles of the Kool-Aid powder is 0.01. We also know that the chemical formula for this powder is
.
We can find the molar weight of this powder by taking the atomic weight from the periodic table and multiply it by the number of atoms in the compound. Let's find the molar weight of the powder:
Find the atomic weight of carbon:
There are 12 C's, so we multiply by 12:
Then we have 22 H's, each having an atomic weight of 1.008g:
And finally there are 11 O's, each with an atomic weight of 15.99g:
Then we add up all of the weights:
-> So we know that the molar weight of the Kool-Aid powder is 342.19g.
Now we can find the amount of grams that we need for the desired solution as follows:
So now we know that we need 0.01 moles of the powder, and we need 3.42g of the powder to make the desired solution.
We know that we want a 0.1M solution in 100 mL, so we can plug those into the equation to find the number of moles required:
So now we know that the number of moles of the Kool-Aid powder is 0.01. We also know that the chemical formula for this powder is
We can find the molar weight of this powder by taking the atomic weight from the periodic table and multiply it by the number of atoms in the compound. Let's find the molar weight of the powder:
Find the atomic weight of carbon:
There are 12 C's, so we multiply by 12:
Then we have 22 H's, each having an atomic weight of 1.008g:
And finally there are 11 O's, each with an atomic weight of 15.99g:
Then we add up all of the weights:
Now we can find the amount of grams that we need for the desired solution as follows:
So now we know that we need 0.01 moles of the powder, and we need 3.42g of the powder to make the desired solution.
Answer:
21.2 moles.
Explanation:
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In this case, for the given chemical reaction, we can see there is a 1:4 mole ratio between tetraphosphorous decaoxide and phosphorous; therefore, the following proportional factor provides the requested moles of phodphorous:
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