Answer:
This is because no energy is being created or destroyed in this system
Explanation:
I think this is correct? I hope it helps.
Answer:-
b. compound b will show an nh absorption while compound a will not.
Explanation:-
For compound B CH3CH2CH2NH2 there are two N-H bonds present. So it will show NH absorption.
But in compound A CH3CH2CH2N(CH3)2 there are no N-H bonds. So it will not show NH absorption.
300cm cubed ( have a great night! )
To find the chemical formula of an ionic compound, the first step is to find the charge of the 2 ions. As given already, the charge of sodium ion is 1+, and carbonate ion has a charge of 2-. We can picture it like that: Sodium ion loses 1 electron and carbonate ion gains 2.
The next step is to find how the 2 ions can lose and gain electrons equally. In this case, since each Na ion only loses 1 electron, it cannot satisfy the need of one carbonate ion, since they need 2, not 1. Therefore, 2 Na ions can cover the need of one carbonate ion. So, the ratio of Na to CO3 ion should be 2:1.
Now just combine the 2 ions, positive one at the front, which makes it NaCO3, make sure you do not add the charge and notice that CO3 is a molecule itself so do not remove the 3. Now because the ratio is 2:1, so the final formula is Na2CO3, no need to add 1 if the ratio is 1.
Your answer should be Na2CO3.
<u>Answer:</u> The vapor pressure of the liquid is 0.293 atm
<u>Explanation:</u>
To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon 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,
= initial pressure which is the pressure at normal boiling point = 1 atm
= pressure of the liquid = ?
= Heat of vaporization = 28.9 kJ/mol = 28900 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = 341.88 K
= final temperature = 305.03 K
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28900J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B341.88%7D-%5Cfrac%7B1%7D%7B305.03%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D-1.228atm%5C%5C%5C%5CP_2%3De%5E%7B-1.228%7D%3D0.293atm)
Hence, the vapor pressure of the liquid is 0.293 atm
