Answer:
16 percent
Explanation:
Just answered the question
The substance would be referred to as a compound substance.
Answer:
9.28
Explanation:
pOH refers to a measure of hydroxide ions concentration. pOH tells about the alkalinity of a solution. If pOH is less than 7 then aqueous solutions are alkaline, acidic if pOH is greater than 7 and neutral if pOH is equal to 7.
Concentration of the hydroxide ions = 1.9 x 10-5 M
pH = 
pOH = 14 - pH
=14 - 4.72 = 9.28
The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.
<h3>How do we calculate pH of weak base?</h3>
pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:
pH = pKb + log([HB⁺]/[B])
pKb = -log(1.8×10⁻⁶) = 5.7
Chemical reaction for C₆H₅NH₂ is:
C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻
Initial: 0.25 0 0
Change: -x x x
Equilibrium: 0.25-x x x
Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]
Kb = x² / 0.25 - x
x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:
1.8×10⁻⁶ = x² / 0.25
x² = (1.8×10⁻⁶)(0.25)
x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]
On putting all these values on the above equation of pH, we get
pH = 5.7 + log(0.67×10⁻³/0.25)
pH = 3.13
Hence pH of the solution is 3.13.
To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361
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The dissociation of both salts NaCl and CaCl₂ are as follows;
NaCl --> Na⁺ + Cl⁻
CaCl₂ --> Ca²⁺ + 2Cl⁻
the molar ratio of NaCl to Cl⁻ is 1:1
therefore number of NaCl moles is equal to number of Cl⁻ ions dissociated from NaCl
then number of Cl⁻ ion moles - 0.233 mol/L x 0.1000 L = 0.0233 mol
molar ratio of CaCl₂ to Cl⁻ ions is 1:2
1 mol of CaCl₂ gives out 2 mol of Cl⁻ ions.
number of CaCl₂ moles - 0.150 mol/L x 0.2500 L = 0.0375 mol
then the number of Cl⁻ ion moles - 0.0375 x 2 = 0.0750
total number of Cl⁻ ion moles = 0.0233 mol + 0.0750 mol = 0.0983 mol
volume of solution - 100.0 + 250.0 = 350.0 mL
concentration of Cl⁻ = 0.0983 mol / 0.3500 L = 0.281 M
concentration of Cl⁻⁻ is 0.281 M