Answer:
True
Explanation:
The arrangement of three groups COOH, CO, OH in the order of reducing boiling point is as follow -
COOH > OH > CO
COOH gets strongly polarised due to the presence of electron withdrawing carboxy group and hence have strong H+ bonds as compared to that of alcohol.
Hence the given statement is true.
Boron have 5 protons and only 4 electrons, which means that it has a positive charge.
(The charge is +1)
Hope this helped!
Answer:
The vapor pressure at 60.21°C is 327 mmHg.
Explanation:
Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.
We need to find vapor pressure at 60.21°C.
The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.
![ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta_%7Bvap%7DH%7D%7BR%7D%28%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%29)
We have given in the question
![P_1=102\ mmHg](https://tex.z-dn.net/?f=P_1%3D102%5C%20mmHg)
![T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol](https://tex.z-dn.net/?f=T_1%3D34.90%5C%C2%B0%5C%20C%3D34.90%2B273.15%3D308.05%5C%20K%5C%5CT_2%3D60.21%5C%C2%B0%5C%20C%3D60.21%2B273.15%3D333.36%5C%20K%5C%5C%5CDelta%7Bvap%7DH%3D39.3%20kJ%2Fmol)
And
is the Universal Gas Constant.
![R=0.008 314 kJ/Kmol](https://tex.z-dn.net/?f=R%3D0.008%20314%20kJ%2FKmol)
![ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7B102%7D%29%3D%5Cfrac%7B39.3%7D%7B0.008314%7D%28%5Cfrac%7B1%7D%7B308.05%7D-%5Cfrac%7B1%7D%7B333.36%7D%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B102%7D%29%3D4726.967%28%5Cfrac%7B333.36-308.05%7D%7B333.36%5Ctimes308.05%7D%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B102%7D%29%3D4726.967%28%5Cfrac%7B25.31%7D%7B333.36%5Ctimes308.05%7D%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B102%7D%29%3D4726.967%28%5Cfrac%7B25.31%7D%7B102691.548%7D%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B102%7D%29%3D1.165)
Taking inverse log both side we get,
![\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg](https://tex.z-dn.net/?f=%5Cfrac%7BP_2%7D%7B102%7D%3De%5E%7B1.165%7D%5C%5C%5C%5CP_2%3D102%5Ctimes%203.20%5C%20mmHg%5C%5CP_2%3D327%5C%20mmHg)
The moles of oxygen produced is calculated as follows
write a balanced reacting equation, that is
Mg(ClO3)2 ---->MgCl2 + 3O2
from the equation above the reacting ratio of Mg(ClO3)2 to O2 is 1: 3 therefore the moles of O2 produced = 5.12 moles x3 =15.36 moles
Answer:
CaC2 + 2H2O > C2H2 + Ca(OH)2
Explanation:
Count the amounts of Ca, C, H, and O, all separately on both sides.
Ca: 1 Ca: 1
C: 2 C:2
H: 2 H: 4
O: 1 O: 2
Now, your goal is to get both sides equal to each other by adding a 2 in front of H2O.