<span>No, I believe this is not an example of a chemical reaction.
What we actually see here is a physical change of the solution. Since we are adding
more water to an aqueous solution which is also made up mostly of water, what
we are simply basically doing is dilution. Since the solution is being diluted,
so definitely the color turned lighter.</span>
First off, With a hypothesis
The mass of the solute required is 250.25 g.
<h3>What is the mass of the solute?</h3>
We know that the number of moles of the solute can be used to obtain the mass of the solute that is required. We can now try to find the mass of the solute that is required.
Concentration of the solution = 0.350M
Volume of the solution = 6.5 L
Number of moles of the solute = 0.350M * 6.5 L
= 2.275 moles
We now have the mass of the solute as;
2.275 moles * 110 g/mol
= 250.25 g
Th measured mass of the solute that we would have to use is 250.25 g.
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Missing parts;
A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?
Answer:
2
Step-by-step explanation:
A. Moles before mixing
<em>Beaker I:
</em>
Moles of H⁺ = 0.100 L × 0.03 mol/1 L
= 3 × 10⁻³ mol
<em>Beaker II:
</em>
Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.
H⁺][OH⁻] = 1 × 10⁻¹⁴ Divide each side by [H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/[H⁺]
[OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)
[OH⁻] = 0.01 mol·L⁻¹
Moles of OH⁻ = 0.100 L × 0.01 mol/1 L
= 1 × 10⁻³ mol
B. Moles after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 3 × 10⁻³ 1 × 10⁻³
C/mol: -1 × 10⁻³ -1 × 10⁻³
E/mol: 2 × 10⁻³ 0
You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.
You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.
C. pH
[H⁺] = (2 × 10⁻³ mol)/(0.200 L)
= 1 × 10⁻² mol·L⁻¹
pH = -log[H⁺
]
= -log(1 × 10⁻²)
= 2
