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Nesterboy [21]
3 years ago
15

What is the concentration of chloride in a solution made with 0.808 grams of CaCl2 and 250.0 ml of water.?

Chemistry
1 answer:
densk [106]3 years ago
8 0

Answer:

0.0584 M

Explanation:

From the question given above, the following data were obtained:

Mass of CaCl₂ = 0.808 g

Volume of water = 250 mL

Concentration of chloride =?

Next, we shall determine the number of mole in 0.808 g of CaCl₂. This can be obtained as follow:

Mass of CaCl₂ = 0.808 g

Molar mass of CaCl₂ = 40 + (35.5 × 2)

= 40 + 71

= 111 g/mol

Mole of CaCl₂ =?

Mole = mass / Molar mass

Mole of CaCl₂ = 0.808 / 111

Mole of CaCl₂ = 0.0073 mole

Next, we shall convert 250 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

250 mL = 250 mL × 1 L / 1000 mL

250 mL = 0.25 L

Next, we shall determine the molarity of CaCl₂. This can be obtained as follow:

Mole of CaCl₂ = 0.0073 mole

Volume of water = 0.25 L

Molarity of CaCl₂ =?

Molarity = mole /Volume

Molarity of CaCl₂ = 0.0073 / 0.25

Molarity of CaCl₂ = 0.0292 M

Finally, we shall determine the concentration of the chloride as illustrated below:

CaCl₂ <=> Ca²⁺ + 2Cl¯

From the equation above,

1 mole of CaCl₂ produced 2 mole of Cl¯.

Therefore, 0.0292 M CaCl₂ will produce = 0.0292 × 2 = 0.0584 M Cl¯.

Thus, the concentration of the chloride ion (Cl¯) in the solution is 0.0584 M

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Answer:

<h2>13.82 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{8.32 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 13.820598...

We have the final answer as

<h3>13.82 moles</h3>

Hope this helps you

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200.00 grams of an organic compound is known to contain 83.884 grams of carbon, 10.486
ololo11 [35]

The empirical formula of a given compound is C6H9ON5.

<u>Explanation</u>:

Step 1: Obtain the mass of each element present in grams

                  Element % = mass in g = m

Carbon = 83.884 grams, Hydrogen = 10.486 grams, Oxygen = 18.640 grams, Nitrogen = 86.99 grams.

Step 2: Determine the number of moles of each type of atom present

                m/atomic mass = Molar amount (M)

Molar amount of carbon = (83.884 1 mol ) / 12 g = 6.99

Molar amount of hydrogen = (10.486  1 mol) / 1 g = 10.49

Molar amount of oxygen = (18.64  1 mol) / 16 g = 1.17

Molar amount of nitrogen = (86.99  1 mol) / 14 g = 6.21

Step 3: Divide the number of moles of each element by the smallest number of moles

            M / least M value = Atomic Ratio (R)

Atomic radius of carbon = 6.99 / 1.17 = 5.9 = 6

Atomic radius of hydrogen = 10.49 / 1.17 = 8.9 = 9

Atomic radius of oxygen = 1.17 / 1.17 = 1

Atomic radius of nitrogen = 6.21 / 1.17 = 5

Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.

            R * whole number = Empirical Formula

The empirical formula of a given compound is C6H9ON5.

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Answer:

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Explanation:

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How many moles of al does 13.49 g of al represent?
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Use equation
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