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
Pollution because it’s one of the reasons for global warming, and global warming causes habitat destruction causing many organisms to become extinct
Answer:
113 g NaCl
Explanation:
The Ideal Gas Law equation is:
PV = nRT
In this equation,
> P = pressure (atm)
> V = volume (L)
> n = number of moles
> R = 8.314 (constant)
> T = temperature (K)
The given values all have to due with the conditions fo F₂. You have been given values for all of the variables but moles F₂. Therefore, to find moles F₂, plug each of the values into the Ideal Gas Law equation and simplify.
(1.50 atm)(15.0 L) = n(8.314)(280. K)
2250 = n(2327.92)
0.967 moles F₂ = n
Using the Ideal Gas Law, we determined that the moles of F₂ is 0.967 moles. Now, to find the mass of NaCl that can react with F₂, you need to (1) convert moles F₂ to moles NaCl (via the mole-to-mole ratio using the reaction coefficients) and then (2) convert moles NaCl to grams NaCl (via molar mass from periodic table). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator).
1 F₂ + 2 NaCl ---> Cl₂ + 2NaF
Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl): 58.44 g/mol
0.967 moles F₂ 2 moles NaCl 58.44 g
---------------------- x ----------------------- x ----------------------- = 113 g NaCl
1 mole F₂ 1 mole NaCl
