To answer this question, we will use the following equation:
<span>ln(P2/P1) = (∆Hvap/R)*((1/T1) - (1/T2))
</span>
Now we examine the givens of the problem and transform to standard units if required:
<span>∆Hvap = 30.5 kJ/mol
</span>R is a constant = <span>8.314 x 10^-3 kJ K^-1 mol^-1
T1 </span><span>= 91 celcius = 91 + 273= 364 Kelvin
</span>T2 = 20 celcius = 20 + 273 = 293 k3lvin
P1 is the standard atmospheric pressure = 760 mmHg
P2 is the value to be calculated
Substitute with these values in the equation:
ln(P2/760) = (30.5 / 8.314 x 10^-3) x ((1 / 364) - (1 / 293))
ln(P2/760) = - 2.4662 (Take the exponential both sides to eliminate the ln)
P2 / 760 = e^(-2.4462) = 0.0866
P2 = 0.0866 x 760 = 65.816 mmHg
Answer:
The molarity of HCl is 0.138 M
Explanation:
The titration reaction is as follows:
2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O
When no more HCl is left, the small excess of Ca(OH)₂ added will cause the pH to rise and the indicator will turn. At this point, the number of moles of Ca(OH)₂ added will be the same as half the number of moles of HCl since 1 mol Ca(OH)₂ reacts with 2 moles HCl. Then:
At the endpoint:
moles Ca(OH)₂ = moles HCl / 2
Knowing the number of moles of Ca(OH)₂ added, we can calculate the number of moles of the acid:
mol Ca(OH)₂ = Volume added * concentration of Ca(OH)₂
mol Ca(OH)₂ = 0.0265 l * 0.130 mol/l = 3.45 x 10⁻³ mol Ca(OH)₂
The number of moles of HCl will be:
mol HCl = 2 * 3.45 x 10⁻³ mol = 6.89 x 10⁻³ mol HCl
This number of moles was present in 50.0 ml, then, in 1000 ml:
mol of HCl in 1000 ml = 6.89 x 10⁻³ mol HCl * (1000ml / 50ml) = 0.138 mol
Then:
Molarity HCl = 0.138 M
Explanation:
Dissolve 93.52g of NaCl in about 400mL of distilled water, then add more water until final volume is 800mL. If starting with a solution or liquid reagent: When diluting more concentrated solutions, decide what volume(V2) and molarity (M2) the final soluble should be.
Explanation:
The specially designed microwave units , for the synthesis , are often very expensive and at times get heated up .
The home microwave is suitable only in some case .
And , new modification done on the home microwave , increases the safety factor .
The chemical reactions which need to be done at very high pressure , are preferred to be carried in specially designed oven .
Hence ,
the home microwave is suitable only for some selected and specific reactions .
Answer :
B
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