Answer:
<u>Mass concentration (g/L) </u><u><em>= 2.49g/L.</em></u>
Explanation:
No. of moles = 
=
= 0.001245 moles
Concentration of KHP (C1) in litres = n/v
=
= 0.062 mol/L
We know that:
=
where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.
Since mole ratio is 1 : 1.
1 mole of NaOH - 40g
0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g
⇒0.0498g of NaOH was used during the titration
<u><em>∴Mass concentration (g/L) = 0.0498g ÷ 0.02L</em></u>
<u><em>= 2.49g/L.</em></u>
Work = force*distance
Work = 670 * 5
Work = 3350 Nm
Answer:
The answer to your question is: 2.20 x 10 ²³ molecules
Explanation:
Data
mass = 45.7 g
molecules of CF₂Cl₂ = ?
Process
1.- Calculate the mass number of CF₂Cl₂
C = 12 F =2 x 19 Cl = 2 x 35.5
total = 12 + 38 + 71
total = 121 g
2.- Use the Avogradro's number to solve the problem
121 g ------------------- 6.023 x 10²³ molecules of CF₂Cl₂
45.7 g --------------- x
x = (45.7 x 6.023 x 10²³) / 121
x = 2.75 x 10²⁵ / 121
x = 2.20 x 10 ²³ molecules