Answer:
0.805 M.
Explanation:
Hello!
In this case, since the molarity of a solution is computing by dividing the moles of solute over the volume of solution in liters (M=n/V), for 15.0 g of potassium chloride (74.55 g/mol) we compute the corresponding moles:
Next, since the volume is 0.2500 in liters, the molarity turns out:
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Answer:
Option D = 0.2 Kj
Explanation:
Given data:
Mass of diethyl ether = 1.0 g
Hvap = 15.7 Kj / mol
Heat absorbed = ?
Solution:
Q = mass × Hvap / molar mass
Q = 1.0 g × 15.7 Kj / mol / 74.12 g/mol
Q = 15.7 Kj / 74.12
Q = 0.212 KJ
Answer:
13.4 (w/w)% of CaCl₂ in the mixture
Explanation:
All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.
To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.
<em>Moles AgCl - Molar mass: 143.32g/mol -:</em>
0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻
<em>Moles CaCl₂:</em>
3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂
<em>Mass CaCl₂ -Molar mass: 110.98g/mol-:</em>
1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture
That means mass percent of CaCl₂ is:
0.207g CaCl₂ / 1.55g * 100 =
<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>
Answer:lmk when u have the answer
Explanation:
Answer:
Explanation:
half life = 4 h
initial concentration = 100
final concentration = 15
Time = t
No of half life x = t / 4
15 = 100 x ( 1/2 )ˣ
.15 = ( 1/2 )ˣ
ln .15 = - x ln2
x = - ln .15 / ln 2
= 1.897 / .693
x = 2.737
x = t / 4
t = 2.737 x 4 = 11 h approx .
