Answer:
184 mL.
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 62.8 mL
Initial concentration (C1) = 167 μM
Final concentration (C2) = 57.0 μM
Final volume (V2) =?
To obtain the final volume of the solution, we shall apply the dilution formula as follow:
C1V1 = C2V2
167 × 62.8 = 57 × V2
10487.6 = 57 × V2
Divide both side by 57
V2 = 10487.6 / 57
V2 = 183.99 ≈ 184 mL
Therefore, chemist must add distilled water to the solution until it reaches a volume of 184 mL.
C) Each second, solid sodium chloride dissolves and solid sodium chloride recrystallizes
All of these properties are known as physical properties.
Physical properties are such that they are measured without changing the composition of the of the matter which is under consideration. For example, the melting point of a substance may be tested without changing its composition. The change is also reversed easily. Physical properties are used to describe and observe matter. The other type of properties are chemical properties, which require the composition of a substance to be changed for them to be measured.
Answer: The concentration of KOH for the final solution is 0.275 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
where,
n = moles of solute
= volume of solution in ml = 150 ml
moles of solute =
Now put all the given values in the formula of molality, we get
According to the dilution law,
where,
= molarity of stock solution = 1.19 M
= volume of stock solution = 15.0 ml
= molarity of diluted solution = ?
= volume of diluted solution = 65.0 ml
Putting in the values we get:
Therefore, the concentration of KOH for the final solution is 0.275 M
Answer:
Explanation:
This is an example of back titration.
You add a known excess of HCl to the CaCO₃. Some of it reacts with the CaCO₃.
You titrate the left-over HCl.
From the difference you can calculate the amount of CaCO₃.
100.09
CaCO₃ + 2HCl ⟶ CaCl₂ + H₂CO₃
m/g: 1.2450
V/mL: 65.00
c/mol·L⁻¹: 0.4984
NaOH + HCl ⟶ NaCl + H₂O
V/mL: 37.15
c/mol·L⁻¹: 0.2065
1. Total moles of HCl
2. Excess moles of HCl
3. Moles of HCl that reacted
n = (32.40 - 7.671) mmol HCl = 24.72 mmol HCl
4. Moles of CaCO₃
5. Mass of Ca