Answer:
The molarity (M) of the following solutions are :
A. M = 0.88 M
B. M = 0.76 M
Explanation:
A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.
Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)
= 27 + 3(16 + 1)
= 27 + 3(17) = 27 + 51
= 78 g/mole
= 78 g/mole
Given mass= 19.2 g/mole


Moles = 0.246

Volume = 280 mL = 0.280 L

Molarity = 0.879 M
Molarity = 0.88 M
B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr
Molar mass of KBr = 119 g/mole
Given mass = 235.9 g

Moles = 1.98
Volume = 2.6 L


Molarity = 0.762 M
Molarity = 0.76 M
One mole is always the same number: 6.02 * 10^ 23.
So, one mole of cars = 6.02 * 10 ^23 cars; one mole of pencils = 6.02 * 10^23 pencils; one mole of atoms = 6.02 * 10^23 atoms; one mole of molecules = 6.02 * 10^23 molecules.
So, all the options are correct: one mole of calcium ions has 6.02 * 10^23 representative particles, such as one mole of calcium nuclei and one of calcium atoms.
Electrolysis and heat can separate a compound into its elements.
I hope this helped you!
Answer:
4.1x10⁻⁵
Explanation:
The dissociation of an acid is a reversible reaction, and, because of that, it has an equilibrium constant, Ka. For a generic acid (HA), the dissociation happens by:
HA ⇄ H⁺ + A⁻
So, if x moles of the acid dissociates, x moles of H⁺ and x moles of A⁻ is formed. the percent of dissociation of the acid is:
% = (dissociated/total)*100%
4.4% = (x/[HA])*100%
But x = [A⁻], so:
[A⁻]/[HA] = 0.044
The pH of the acid can be calcualted by the Handersson-Halsebach equation:
pH = pKa + log[A⁻]/[HA]
3.03 = pKa + log 0.044
pKa = 3.03 - log 0.044
pKa = 4.39
pKa = -logKa
logKa = -pKa
Ka = 
Ka = 
Ka = 4.1x10⁻⁵