Answer:
0.143 g of KCl.
Explanation:
Equation of the reaction:
AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)
Molar concentration = mass/volume
= 0.16 * 0.012
= 0.00192 mol AgNO3.
By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.
Number of moles of KCl = 0.00192 mol.
Molar mass of KCl = 39 + 35.5
= 74.5 g/mol
Mass = molar mass * number of moles
= 74.5 * 0.00192
= 0.143 g of KCl.
Answer:
I have no clue what the question is
The fraction of the original amount remaining is closest to 1/128
<h3>Determination of the number of half-lives</h3>
- Half-life (t½) = 4 days
- Time (t) = 4 weeks = 4 × 7 = 28 days
- Number of half-lives (n) =?
n = t / t½
n = 28 / 4
n = 7
<h3>How to determine the amount remaining </h3>
- Original amount (N₀) = 100 g
- Number of half-lives (n) = 7
- Amount remaining (N)=?
N = N₀ / 2ⁿ
N = 100 / 2⁷
N = 0.78125 g
<h3>How to determine the fraction remaining </h3>
- Original amount (N₀) = 100 g
- Amount remaining (N)= 0.78125 g
Fraction remaining = N / N₀
Fraction remaining = 0.78125 / 100
Fraction remaining = 1/128
Learn more about half life:
brainly.com/question/26374513
Nolur acil lütfen yalvarırım sana da hayırlı
Answer: the percent by mass of potassium nitrate is 13.3%
Explanation:
<u>1) Data:</u>
- Mass of solute, m₁ = 45.0 g
- Volume of solvent, V₂ = 295 ml
- density of water, d₂ = 0.997 g/ml
<u>2) Formulae</u>:
- Percent by mass, % = (mass of solute / mass of solution) × 100
- Density, d = mass / volume
<u>3) Solution</u>:
- d = mass / volume ⇒ m₂ = d₂ × V₂ = 0.997 g/ml × 295 ml = 294. g
- mass of solution = m₁ + m₂ = 45.0g + 294. g = 339. g
- % = (45.0 g / 339. g) × 100 = 13.3 %
- The answer has to be reported with 3 significant figures.
