Answer:
THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.
Explanation:
Mass of the unknown substance = 0.50 g
Freezing point of the solution = 3.9 °C
Freezing point of pure benzene = 5.5 °C
Freezing point dissociation constant Kf = 5.12°C/m
First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.
Change in temperature = 5.5 -3.9 = 1.6 °C
Next is to calculate the number of moles or molarity of the compound that dissolved.
Using the formula:
Δt = i Kf m
Assume i = 1
So,
1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene
x = 1.6 * 0.008 / 5.12
x = 0.0128 / 5.12
x = 0.0025 moles.
Next is to calculate the molar mass using the formula, molarity = mass / molar mass
Molar mass = mass / molarity
Molar mass = 0.50 g /0.0025
Molar mass = 200 g/mol
Hence, the molar mass of the unknown compound is 200 g/mol
Answer:
A(g)+B(g)⟶C(g) 2A(g)+2B(g)⟶5C(g)
A(s)+B(s)⟶C(g) 2A(g)+2B(g)⟶3C(g):
Word equation: magnesium chloride + bromine -> magnesium bromide + chlorine
balance equation: MgCl2 + 2Br -> MgBr2 + 2Cl
hydrogen is the lightest element in the periodic table
Answer: a) pH = 13.00 : basic
b)
: basic
c) pOH = 5.00 : basic
d)
: acidic
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.
a) pH = 13.00
As pH is more than 7, the solution is basic.
b) ![[H_3O^+]=1.0\times 10^{-12}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D1.0%5Ctimes%2010%5E%7B-12%7D)
Putting in the values:
![pH=-\log[1.0\times 10^{-12}]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5B1.0%5Ctimes%2010%5E%7B-12%7D%5D)

As pH is more than 7, the solution is basic.
c) pOH = 5.00


As pH is more than 7, the solution is basic.
d) ![[OH^-]=1.0\times 10^{-9}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.0%5Ctimes%2010%5E%7B-9%7D)
Putting in the values:
![pOH=-\log[1.0\times 10^{-9}]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B1.0%5Ctimes%2010%5E%7B-9%7D%5D)



As pH is less than 7, the solution is acidic.