<u>Given:</u>
Volume of HCl = 150 ml
Molarity of HCl = 0.10 M
<u>To determine:</u>
The # moles of HCl
<u>Explanation:</u>
The molarity of a solution is the number of moles of a solute dissolved in a given volume
In this case:
Molarity of HCl = moles of HCl/volume of the solution
moles of HCl = Molarity * volume = 0.10 moles.L-1 * 0.150 L = 0.015 moles
Ans: A)
Moles of HCl is 0.015
When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
Answer:
Molecular formula = C₄H₆As₆Cu₄O₁₆
Explanation:
Given data:
Empirical formula = C₂H₃As₃Cu₂O₈
Molar mass of compound = 1013 g/mol
Molecular formula = ?
Solution:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass of C₂H₃As₃Cu₂O₈ is 506.897 g/mol
by putting values.
n = 1013 / 506.897
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 (C₂H₃As₃Cu₂O₈)
Molecular formula = C₄H₆As₆Cu₄O₁₆
Answer:
B
Explanation:
I looked it up and found the answer lol
Hydrochloric acid and sodium hydroxide
HCl + NaOh ----> NaCl + H2O