From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.
From the information we have;
Volume of the damp air = 1 L
Pressure of the damp air = 741.0 torr or 0.975 atm
Temperature of the gas = 20 oC + 273 = 293 K
R = 0.082 atm LK-1mol-1
Number of moles = ?
n =PV/RT
n = 0.975 × 1/0.082 × 293
n = 0.041 moles
Volume of water vapor = 1 L
Temperature of water = -10 oC + 273 = 263 K
Pressure of the gas = 607.1 torr or 0.799 atm
R = 0.082 atm LK-1mol-1
n= PV/RT
n = 0.799 × 1/ 0.082 × 263
n = 0.037 moles
Number of moles of water = 0.041 moles - 0.037 moles = 0.004 moles
If 1 mole = 6.02 × 10^23 molecules
0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole
= 2.41 × 10^21 molecules
Learn more: brainly.com/question/2510654
Answer:
18,1 mL of a 0,304M HCl solution.
Explanation:
The neutralization reaction of Ba(OH)₂ with HCl is:
2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O
The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:
= 2,7531x10⁻³moles of Ba(OH)₂
By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:
= 5,5062x10⁻³moles of HCl.
The volume of a 0,304M HCl solution for a complete neutralization is:
= 0,0181L≡18,1mL
I hope it helps!
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