A catalyst is when a chemical reaction occurs faster than normal.
The system is unaffected during a catalyst because both forward and reverse reactions are affected, meaning that quilibrium will occur faster nothing will change.
Hope it helped,
BioTeacher101
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<span>if we assume volume NaCl + volume H2O = volume H2O.. i.e.. NaCl does not effect volume </span>
<span>therefore.. the units of.. </span>
<span>.. M = moles NaCl / L solution ≈ moles NaCl / L H2O </span>
<span>.. density = grams NaCl / L solution ≈ grams NaCl / L H2O </span>
<span>again.. that is our assumption </span>
<span>so we can readily see that </span>
<span>.. M = (1 mol NaCl / ___g NaCl) x (__g NaCl / L H2O) + 0 </span>
<span>ie.. </span>
<span>.. M = (1 mol NaCl / 58.5g NaCl) x density solution + 0 </span>
<span>so.. we would expect.. </span>
<span>.. m = 0.01709 mol / g </span>
<span>.. b = 0 </span>
Answer:
Water will boil at 
.
Explanation:
According to clausius-clapeyron equation for liquid-vapor equilibrium:
![ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_%7B2%7D%7D%7BP_%7B1%7D%7D%29%3D%5Cfrac%7B-%5CDelta%20H_%7Bvap%7D%5E%7B0%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7BT_%7B1%7D%7D%5D)
where, 
and 
are vapor pressures of liquid at 
(in kelvin) and 
(in kelvin) temperatures respectively.
Here, = 760.0 mm Hg, = 373 K, = 314.0 mm Hg
Plug-in all the given values in the above equation:
![ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B314.0%7D%7B760.0%7D%29%3D%5Cfrac%7B-40.7%5Ctimes%2010%5E%7B3%7D%5Cfrac%7BJ%7D%7Bmol%7D%7D%7B8.314%5Cfrac%7BJ%7D%7Bmol.K%7D%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_%7B2%7D%7D-%5Cfrac%7B1%7D%7B373K%7D%5D)
or, 
So, 
Hence, at base camp, water will boil at
At stp the volume is 22.4 L .
hope this helps!
