Answer:
Mass of solute = 0.0036 g
Explanation:
Given data:
Concentration of Cl⁻ = 15.0 ppm
Volume of water = 240 mL
Mass of Cl⁻ present = ?
Solution:
1 mL = 1 g
240 mL = 240 g
Formula:
ppm = mass of solute / mass of sample ×1,000,000
by putting values,
15.0 ppm = (mass of solute / 240 g) ×1,000,000
Mass of solute = 15.0 ppm × 240 g / 1,000,000
Mass of solute = 0.0036 g
Missing question:
Chemical reaction: H₂ <span>+ 2ICl → 2HCl + I</span>₂.
t₁ = 5 s.
t₂ = 15 s.
c₁ = 1,11 M = 1,11 mol/L.
c₂ = 1,83 mol/L.
rate of formation = Δc ÷ Δt.
rate of formation = (c₂ - c₁) ÷ (t₂ - t₁).
rate of formation = (1,83 mol/L - 1,11 mol/L) ÷ (15 s - 5 s).
rate of formation = 0,72 mol/L ÷ 10 s.
rate of formation = 0,072 mol/L·s.
Answer:
1) THE AGE OF THE SAMPLE 2) URANIUM- LEAD DATING
Explanation:
Explanation:
o2- Due to lowest Zeff.
I had that question and my answer was this. because you did not put options.
In an average mass, each entry has equal weight. In a weighted average, we multiply each entry by a number representing its relative importance.
Assume that your class consists of 15 girls and 5 boys. Each girl has a mass of 54 kg, and each boy has a mass of 62 kg.
<em>Average mass</em> = (girl + boy)/2 = (54 kg + 62 kg)/2 = <em>58 kg</em>
<em>Weighted average (Method 1)
</em>
Use the <em>numbers of each</em> gender (15 girls + 5 boys)
,
Weighted average = (15×54 kg + 5×62 kg)/20 = (810 kg + 310 kg)/20
= 1120 kg/20 = <em>56 kg</em>.
If you put all the students on one giant balance, their total mass would be
1120 kg and the average mass of a student would be <em>56 kg.
</em>
<em>Weighted average (Method 2)
</em>
Use the <em>relative percentages</em> of each gender (75 % girls and 25 % boys).
Weighted average = 0.75×54 kg + 0.25×62 kg = 40.5 kg + 15.5 kg = <em>56 kg</em>
Each girl contributes 40.5 kg and each boy contributes 15.5 kg to the <em>weighted average</em> mass of a student.
