The correct answer for the question that is being presented above is this one:
Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m
<span>delta Tf = Kfm Kf H2O = 1.86 degrees C/m
</span>
We need to know the formula for Molality.
molality = mol solute / kg solvent
<span>We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. </span>
<span>25 g NaCl / 58.5 g/mol = 0.427 mol </span>
<span>Then, use the formula for molality. </span>
<span>molality = mol solute / kg solvent </span>
<span>= 0.427 / 1 </span>
<span>= 0.427 m </span>
<span>Use now the formula to get the boiling point.</span>
<span>delta Tb = Kbm </span>
<span>= (0.52)(0.427) </span>
<span>= 0.22C </span>
For equal moles of gas, temperature can be calculated from ideal gas equation as follows:
P×V=n×R×T ...... (1)
Initial volume, temperature and pressure of gas is 3.25 L, 297.5 K and 2.4 atm respectively.
2.4 atm ×3.25 L=n×R×297.5 K
Rearranging,
n\times R=0.0262 atm L/K
Similarly at final pressure and volume from equation (1),
1.5 atm ×4.25 L=n×R×T
Putting the value of n×R in above equation,
1.5 atm ×4.25 L=0.0262 (atm L/K)×T
Thus, T=243.32 K
Answer:
Explanation:
The formula relating the mass m of a sample and the heat q to vaporize it is
q = mL, where L is the latent heat of vaporization.

Answer:
87.15%
Explanation:
To find percent yield, we can use this simple equation

Where "Actual" is the amount in grams actually collected from the reaction, and "Theoretical" is, well, the theoretical amount that should have been produced.
They give us these values, so to find the percent yield, just plug the numbers in.

So, the percent yield is 87.15%
An easy trick to remember how to do this is just to divide the smaller number by the bigger number and move the decimal back two places. If you have a percent yield greater than 100%, something is wrong in the reaction.
Answer:
The answer to this question has been described in details on the screenshots attached to this question.
Thanks. Hope it helps