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algol [13]
2 years ago
7

What is the concentration of a solution with a volume of 660 mL that contains 33.4 grams of aluminum acetate?

Chemistry
2 answers:
Damm [24]2 years ago
4 0

Answer:

The concentration is 50.6 g/L

Explanation:

pa BRAINLYES PO TY

Olin [163]2 years ago
3 0
The concentration is 50.6 g/L
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A student was given the task of titrating a 20.mL sample of 0.10MHCl(aq) with 0.10MNaOH(aq) . The HCl(aq) was placed in an Erlen
gregori [183]

Answer:

Here's what I get  

Step-by-step Explanation

(a) Effect of dilution

There will be no effect on the volume of NaOH needed.

The amount of HCl will be halved, so the amount of NaOH will be halved.

However, the concentration of NaOH is also halved, so you will need twice the volume.

You will be back to the same volume as before dilution.

(b) Net ionic equation

Molecular: HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

Ionic: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) ⟶ Na⁺(aq) + Cl⁻(aq) + H₂O(l)

Net ionic: H⁺(aq) + OH⁻(aq) ⟶ H₂O(l)

(c) Proton acceptor

H⁺ is the proton. OH⁻ accepts the proton and forms water.

(d) Moles of HCl

\text{Moles of HCl} = \text{20. mL HCl} \times \dfrac{\text{0.10 mmol HCl}}{\text{1 mL HCl}} = \text{2.0 mmol HCl} =  \textbf{0.0020 mol HCl}

(e) Equivalence point

The equivalence point is the point at which the titration curve intersects the pH 7 line.

(f) Schematic representation

Assume the box for 0.10 mol·L⁻¹ HCl contains four black dots (H⁺) and four open circles (Cl⁻).

The 0.20 mol·L⁻¹ solution is twice as concentrated.

It will contain eight black dots and eight open circles.

5 0
3 years ago
A chemist prepares a solution of silver nitrate by measuring out of silver nitrate into a volumetric flask and filling the flask
Snezhnost [94]

Amount of silver nitrate taken = 269.μmol AgNO_{3}

Volume of the solution = 300. mL

Concentration of a solution is generally expressed in terms of molarity. Molarity is defined as the moles of a substance present per liter of the solution.

Molarity = \frac{Moles of solute}{Volume of solution(L)}

We want the concentration in millimoles/L.

Converting μmol to millimol solute:

269.μmol * \frac{1 millimol}{1000 micromol} = 0.269 millimol

Volume from mL to L: 300. mL * \frac{1 L}{1000 mL} = 0.300 L

Therefore concentration of the chemist's solution = \frac{0.269 millimol}{0.300 L} =  0.897 \frac{millimol}{L}

8 0
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