Answer:
the concentration of the reactants
the temperature in heating
Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %
Answer:
They have the same number of valence electrons (5) so the body assumes they are the same element.
Answer:
C
Explanation:
I just aswered this question and i got it right
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
