If someone consumes 22 g of sodium chloride per day, what mass (in grams) of sodium does that person consume? Sodium chloride is
1 answer:
Answer:
Mass of sodium, the person daily consume = 8.58 g
Explanation:
Given that:-
sodium chloride is 39% sodium by mass which means that:-
100 g of sodium chloride contains 39 g of sodium
Also,
1 g of sodium chloride contains 0.39 g of sodium
So, Given mass = 22 g
22 g of sodium chloride contains g of sodium
<u>Mass of sodium, the person daily consume = 8.58 g</u>
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Answer:
5.42g, 71.77%
Explanation:
First, we have to write out the balanced chemical equation. The unbalanced equation can be written as “SO2+O2 -> SO3” and to balance it, we can see that having two mols of SO2 and two mols of SO3 will make each side have the same amount of mols per element on each side. So the balanced chemical equation is “2SO2 + O2 -> 2SO3”
Now, we want to solve for the theoretical yield in grams of SO3. To do this, we have to use dimensional analysis. We convert g SO2 into mols SO2 using the molar mass of the elements. Then we convert mols of SO2 into mols of SO3 using the balanced equation. Once we’ve done that, we can convert mols of SO3 into grams of SO3.
You should know how to look up the molar mass of elements on the periodic table by now. Find the masses and set up the terms so they cancel like so:
Doing the math, we get 5.42g so3 as the theoretical yield. This is the most amount that you could ever get if the world was a perfect place. But alas, it isn’t and mistakes are gonna happen, so the number is going to be less than that. So the best we can do, is to figure out the percent yield that we got.
In a lab scenario, this was calculated to be 3.89 g as stated by the problem. The percent composition formula is
and plugging the numbers into it, we get:
make sure to follow the decimal/significant figure rules of your instructor, but only round at the end. My professor didn't care too much thankfully, but some professors do
Answer:
2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.
12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution
Explanation:
First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:
- Ca: 40 g/mole
- F: 19 g/mole
So the molar mass of CaF₂ is:
CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?
moles=2.05*10⁻⁵
<u><em>2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.</em></u>
Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?
mass of CaF₂= 1000 g
Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?
moles=12.82
<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>