I think it’s 1 mole
If Iam not mistaken
The formula of the ppt. formed is PbSo4 , which is inslouble.
First, consider the steps to heat the sample from 209 K to 367K.
1) Heating in liquid state from 209 K to 239.82 K
2) Vaporaizing at 239.82 K
3) Heating in gaseous state from 239.82 K to 367 K.
Second, calculate the amount of heat required for each step.
1) Liquid heating
Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol
=> number of moles = 12.62 g / 17 g/mol = 0.742 mol
Heat1 = #moles * heat capacity * ΔT
Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J
2) Vaporization
Heat2 = # moles * H vap
Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J
3) Vapor heating
Heat3 = #moles * heat capacity * ΔT
Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J
Third, add up the heats for every steps:
Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J
Fourth, divide the total heat by the heat rate:
Time = 22,466.3 J / (6000.0 J/min) = 3.7 min
Answer: 3.7 min
The pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
<h3>What is pH? </h3>
pH is defined as the concentration of the hydrogen bond which is released or gained by the species in the solution which depicts the acidity and basicity of the solution.
<h3>What is pOH? </h3>
pOH is defined as the concentration of the hydronium ion present in solution.
pOH value is inversely proportional to the value of pH.
pH value increases, pOH value decreases and vice versa.
Given,
Total H+ ions = 2.95 ×10^(-12)M
<h3>Calculation of pH</h3>
pH = -log[H+]
By substituting the value of H+ ion in given equation
= log(2.95× 10^(-12) )
= 13.5
Thus we find that the pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
learn more about pH:
brainly.com/question/12942138
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Answer:
Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.
Explanation:
If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.
