How many molecules are in 2L of water vapor at STP?
1 answer:
Answer:
5.4 x 10²²molecules
Explanation:
Given parameters = 2L
Unknown; number of molecules
Condition of the water vapor is at standard Temperature and Pressure
Solution
To solve this problem, we simply find the number of moles the given volume would yield at STP using the relationship below:
Number of moles =
Number of moles = = 0.089mole
To find the number of molecules in the solved mole of the water vapor:
we use the avogadro's constant to evaluate.
1 mole of a water vapor = 6.02 x 10²³ molecules
0.089 mole of water vapor = 0.54 x 10²³molecules
Number of molecules of water is 5.4 x 10²²molecules
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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>