Molarity's formula is known as: Molarity(M)=moles of solute/liters solution.
In this case we are already given moles and liters so you just have to plug the numbers into the equation.
0.400 mol HCL/9.79L solution=0.040858M
If you were to use scientific notation, the answer will be: 4.1*10^-2, but otherwise, you can just use the decimals above and round appropriately as you see fit.
Answer: The value of 
for chloroform is 
when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Explanation:
Given: Moles of solute = 0.793 mol
Mass of solvent = 0.758
As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.
Now, the values of 
is calculated as follows.
where,
i = Van't Hoff factor = 1 (for chloroform)
m = molality
= molal boiling point elevation constant
Substitute the values into above formula as follows.
Thus, we can conclude that the value of for chloroform is when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Answer:
K(48.5°C) = 1.017 E-8 s-1
Explanation:
- CH3Cl + H2O → CH3OH + HCl
at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1
at T2 = 48.5°C (321.5 K) ⇒ K2 = ?
Arrhenius eq:
- K(T) = A e∧(-Ea/RT)
- Ln K = Ln(A) - [(Ea/R)(1/T)]
∴ A: frecuency factor
∴ R = 8.314 E-3 KJ/K.mol
⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)
⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)
(1)/(2):
⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)
⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)
⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)
⇒ Ln (K1/K2) = - 3.422
⇒ K1/K2 = e∧(-3.422)
⇒ (3.32 E-10 s-1)/K2 = 0.0326
⇒ K2 = (3.32 E-10 s-1)/0.0326
⇒ K2 = 1.017 E-8 s-1
Answer:
because it is type of tht a solid
