Answer:
Bones are the correct answer.
Answer:
(a) 
(b) 
Explanation:
Hello,
(a) In this case, for the given chemical reaction, the law of mass action becomes:
![Kc=\frac{[C6H5CHO][H2]}{[C6H5CH2OH]}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC6H5CHO%5D%5BH2%5D%7D%7B%5BC6H5CH2OH%5D%7D)
In such a way, as 1.20 g of benzyl alcohol are placed into a 2.00-L vessel, the initial concentration is:
![[C6H5CH2OH]_0=\frac{1.20g*\frac{1mol}{108.14g} }{2.00L} =5.55x10^{-3}M](https://tex.z-dn.net/?f=%5BC6H5CH2OH%5D_0%3D%5Cfrac%7B1.20g%2A%5Cfrac%7B1mol%7D%7B108.14g%7D%20%7D%7B2.00L%7D%20%3D5.55x10%5E%7B-3%7DM)
Hence, by writing the law of mass action in terms of the change 
due to equilibrium:
Solving for by using a quadratic equation one obtains:
Thus, the equilibrium concentration of benzyl alcohol is computed:
![[C6H5CH2OH]_{eq}=5.55x10^{-3}M-5.50x10^{-3}M=5x10^{-5}M](https://tex.z-dn.net/?f=%5BC6H5CH2OH%5D_%7Beq%7D%3D5.55x10%5E%7B-3%7DM-5.50x10%5E%7B-3%7DM%3D5x10%5E%7B-5%7DM)
With that concentration the partial pressure results:
![p_{C6H5CH2OH}=[C6H5CH2OH]_{eq}RT =5x10^{-5}\frac{mol}{L} *0.082\frac{atm*L}{mol*K}*523K \\p_{C6H5CH2OH}=2.14x10^{-3}atm](https://tex.z-dn.net/?f=p_%7BC6H5CH2OH%7D%3D%5BC6H5CH2OH%5D_%7Beq%7DRT%20%3D5x10%5E%7B-5%7D%5Cfrac%7Bmol%7D%7BL%7D%20%2A0.082%5Cfrac%7Batm%2AL%7D%7Bmol%2AK%7D%2A523K%20%5C%5Cp_%7BC6H5CH2OH%7D%3D2.14x10%5E%7B-3%7Datm)
(b) Now, the fraction of benzyl alcohol that is dissociated relates its equilibrium concentration with its initial concentration:
Best regards.
Its temperature will rise until reaching 0° C Its temperature will remain at 0° C until all the ice melts
Br
Se
As
Ge
Ga
Rb
Elements in the top right corner of the periodic table have the highest electronegativity. Elements on the right side have a higher electronegativity than those on the left, same with the ones on the top in comparison to those on the bottom.
The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is : - 0.69°C
<h3>Determine the freezing point of the solution </h3>
First step : Calculate the molality of NaCl
molality = ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )
= 0.186 mol/kg
Next step : Calculate freezing point depression temperature
T = 2 * 0.186 * kf
where : kf = 1.86°c.kg/mole
Hence; T = 2 * 0.186 * 1.86 = 0.69°C
Freezing point of the solution
Freezing temperature of solvent - freezing point depression temperature
0°C - 0.69°C = - 0.69°C
Hence the Freezing temperature of the solution is - 0.69°C
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