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
Nitrous acid<span> dissociates as follows:
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HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq)
According to the equation, an acid constant has the following form:
Ka = [H⁺] × [NO₂⁻ ] / [HNO₂]
From pH, we can calculate the concentration of H⁺ and NO₂⁻:
[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]
Now, the acid constant can be calculated:
Ka = 0.00234 x 0.00234 / 0.015 = 3.66 x 10⁻⁴
And finally,
pKa = -log Ka = 3.44
Answer: 3
Explanation: I took the k12 quiz
Using the Fundamental Equation of Calorimetry, we have:
If you notice any mistake with my english, please know me, because I am not native.
