Mixtures can be separated by filtration and distillation. Both methods take advantage of_______. A. differences in the chemical
1 answer:
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Answer:
empirical formula:
2 g H
32.7 g S
65.3 g O
Explanation:
Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:
- 2.00% * 100g = 2 g H
- 32.7% * 100g = 32.7 g S
- 65.3% * 100g = 65.3 g O
Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:
- the molar mass of H is 1.01 g/mol
- the molar mass of S is 32.06 g/mol
- the molar mass of O is 16 g/mol
2 g H ÷ 1.01 g/mol = 1.98 mol H
32.7 g S ÷ 32.06 g/mol = 1.02 mol S
65.3 g O ÷ 16 g/mol = 4.08 mol O
Our ratio of H : S : O is now:
1.98 mol : 1.02 mol : 4.08 mol
Divide them all by the smallest number, which is 1.02:
1.98/1.02 : 1.02/1.02 : 4.08/1.02
1.94 : 1 : 4
1.94 ≈ 2
So:
2 : 1 : 4
Thus, the empirical formula is: .
Answer:
D. The final concentration of NO3– is 0.821 M.
Explanation:
Considering:
Or,
Given :
For potassium iodide :
Molarity = 1.60 M
Volume = 300.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 300.0×10⁻³ L
Thus, moles of potassium iodide :
<u>Moles of potassium iodide = 0.48 moles </u>
For lead(II) nitrate :
Molarity = 1.20 M
Volume = 285 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 285×10⁻³ L
Thus, moles of lead(II) nitrate :
<u>Moles of lead(II) nitrate = 0.342 moles </u>
According to the given reaction:
2 moles of potassium iodide react with 1 mole of lead(II) nitrate
1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate
0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate
Moles of lead(II) nitrate = 0.24 moles
Available moles of lead(II) nitrate = 0.342 moles
<u>Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.</u>
Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)
Left over moles = 0.342 - 0.24 moles = 0.102 moles
Total volume = 300 + 285 mL = 585 mL = 0.585 L
<u>So, Concentration = 0.102/0.585 M = 1.174 M</u>
<u>Statement A is correct.</u>
The formation of the product is governed by the limiting reagent. So,
2 moles of potassium iodide gives 1 mole of lead(II) iodide
1 mole of potassium iodide gives 1/2 mole of lead(II) iodide
0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide
Mole of lead(II) iodide = 0.24 moles
Molar mass of lead(II) iodide = 461.01 g/mol
<u>Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g </u>
<u>Statement B is correct.</u>
Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,
2 moles of Potassium iodide on reaction forms 2 moles of potassium ion
0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion
Total volume = 300 + 285 mL = 585 mL = 0.585 L
<u>So, Concentration = 0.48/0.585 M = 0.821 M</u>
<u>Statement C is correct.</u>
Nitrate ions are furnished by lead(II) nitrate . So,
1 mole of lead(II) nitrate produces 2 moles of nitrate ions
0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions
Moles of nitrate ions = 0.684 moles
<u>So, Concentration = 0.684/0.585 M = 1.169 M</u>
<u>Statement D is incorrect.</u>