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Fittoniya [83]
3 years ago
10

What is the molarity (M) of chloride ions in a solution prepared by mixing 155 ml of 0.276 M calcium chloride with 384 ml of 0.4

71 M aluminum chloride?
Chemistry
1 answer:
sesenic [268]3 years ago
6 0

Answer: The concentration of Cl^- ions in the resulting solution is 1.16 M.

Explanation:

To calculate the molarity of the solution after mixing 2 solutions, we use the equation:

M=\frac{n_1M_1V_1+n_2M_2V_2}{V_1+V_2}

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of the CaCl_2

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of the AlCl_3

We are given:

n_1=2\\M_1=0.276M\\V_1=155mL\\n_2=3\\M_2=0.471M\\V_2=384mL  

Putting all the values in above equation, we get

M=\frac{(2\times 0.276\times 155)+(3\times 0.471\times 384)}{155+384}\\\\M=1.16M

The concentration of Cl^- ions in the resulting solution will be same as the molarity of solution which is 1.16 M.

Hence, the concentration of Cl^- ions in the resulting solution is 1.16 M.

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How many milliliters of an aqueous solution of 0.227 M sodium carbonate is needed to obtain 3.55 grams of the salt?
Marianna [84]

Answer:

147.6 mL .

Explanation:

sodium carbonate , Na₂CO₃

Molecular weight = 106

3.55 gm of sodium carbonate = 3.55 / 106

= .0335 moles

Let the volume of litre required = V

V litre of .227 M solution will contain

V x .227 moles of sodium carbonate . So

V x .227 = .0335

V = .1476 L

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Explanation:

be2+ stable

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Br- stable

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How many atoms are present in a sample of Potassium (K) weighing 33.49g?
madam [21]

Answer:

5.158 × 10²³ atoms K

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
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Explanation:

<u>Step 1: Define</u>

33.49 g K

<u>Step 2: Identify Conversions</u>

Avogadro's Number

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<u>Step 3: Convert</u>

<u />33.49 \ g \ K(\frac{1 \ mol \ K}{39.10 \ g \ K} )(\frac{6.022 \cdot 10^{23} \ atoms \ K}{1 \ mol \ K} ) = 5.15797 × 10²³ atoms K

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig figs and round.</em>

5.15797 × 10²³ atoms K ≈ 5.158 × 10²³ atoms K

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