Answer : The normal boiling point of ethanol will be, 
or 
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of ethanol at 
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the normal boiling point of ethanol will be, or
Answer:
0.0295M
Explanation:
As you can see, in the mixture you have KSCN and other compounds. The KSCN in solution is dissolved in K⁺ ions and SCN⁻ ions. That means initial concentration of SCN⁻ ions is the same of KSCN, 0.0800M.
You are adding 35.0mL of this solution and the total volume of the mixture is 20.0mL + 35.0mL + 40.0mL = 95.0mL.
That means you are diluting your solution 95.0mL / 35.0mL = 2.714 times.
And the concentration of SCN⁻ is:
0.0800M / 2.714 =
<h3>0.0295M </h3>
Answer:
if i remember correctly i beleive its A 1.8 x 10^24
but im not for sure also i think you forgot the 24
Explanation:
2 cm east is the answer because you go east 5.5 cm and then go back in a sense 3.5 so its basically 5.5-3.5 because its backwards while still facing eat
