Answer : The value of
is 28.97 kJ/mol
Explanation :
To calculate
of the reaction, we use clausius claypron equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= vapor pressure at temperature
= 462.7 mmHg
= vapor pressure at temperature
= 140.5 mmHg
= Enthalpy of vaporization = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![-21.0^oC=[-21.0+273]K=252K](https://tex.z-dn.net/?f=-21.0%5EoC%3D%5B-21.0%2B273%5DK%3D252K)
= final temperature = ![45^oC=[-41.0+273]K=232K](https://tex.z-dn.net/?f=45%5EoC%3D%5B-41.0%2B273%5DK%3D232K)
Putting values in above equation, we get:
![\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B140.5mmHg%7D%7B462.7mmHg%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B252%7D-%5Cfrac%7B1%7D%7B232%7D%5D%5C%5C%5C%5C%5CDelta%20H_%7Bvap%7D%3D28966.6J%2Fmol%3D28.97kJ%2Fmol)
Therefore, the value of
is 28.97 kJ/mol
Answer:
c
Explanation:
because nothing is changing, so what are you gonna track with the line graph
Add 4H2 in the reactant side, that will give you 4H2O in the product side.
Answer:
Na₂CO₃•H₂O
Explanation:
After it is heated, the remaining mass is the mass of sodium carbonate.
30.2 g Na₂CO₃
Mass is conserved, so the difference is the mass of the water:
35.4 g − 30.2 g = 5.2 g H₂O
Convert masses to moles:
30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃
5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O
Normalize by dividing by the smallest:
0.285 / 0.285 = 1.00 mol Na₂CO₃
0.289 / 0.285 = 1.01 mol H₂O
The ratio is approximately 1:1. So the formula of the hydrate is Na₂CO₃•H₂O.
Answer:
See explanation
Explanation:
For this question, we have to remember the effect of an atom with high <u>electronegativity</u> as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an <u>inductive effect</u>. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be <u>weaker</u> and the compound will be more acid (because is easier to produce the hydronium ion
).
With this in mind, for A in the last compound, we have <u>2 Br atoms</u> near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have <u>more acidity</u>. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.
In B, the difference between the molecules is the <u>position</u> of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a <u>higher inductive effect</u> and more <u>acidity</u>.
See figure 1
I hope it helps!