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Dafna1 [17]
3 years ago
5

2. What is the molarity of 100 mL of a 3.0% H2O2 (mass/volume) solution? What is the molarity of 100 mL of a 2.25% H2O2 solution

?
Chemistry
1 answer:
KatRina [158]3 years ago
7 0

The 3% mass/volume H₂O₂ means 3 g of H₂O₂ in 100 ml of water.

Now, Molarity (M) = No. of moles of H₂O₂ / Volume of solution in liter

No. of moles of H₂O₂ = Mass / Molar mass = 3 g / 34 g/mol = 0.088 mol

So, molarity = 0.088 × 1000 ml / 100 ml = 0.88 M

In case of 2.25 % H₂O₂,

No of moles = 2.25 g / 34 g/mol = 0.066 mol

Molarity = 0.066 mol / 0.100 L = 0.66 M.

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By what factor does [h+ ] change for each ph change? (a) 3.20 units
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When a change in PH = 10^-ΔPH
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3 0
3 years ago
How many liters of a 1.5 M solution can you make if you have .50 mol of KCl?
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Answer:

The answer to your question is 0.33 liters

Explanation:

Data

Volume = ?

Molarity = 1.5 M

number of moles = 0.5

Formula

Molarity = \frac{number of moles }{volume}

Solve for V

Volume = \frac{number of moles}{molarity}

Substitution

Volume = \frac{0.5}{1.5}

Simplification and result

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7 0
3 years ago
Suppose a large atom bonds with a small
Zielflug [23.3K]
A large atom means that the radius would be large, meaning that the effective nuclear charge is low, therefore a lower electronegativity based on the periodic table. A smaller atom would mean the opposite, therefore a higher electronegativity. This combination would mean that the new molecule is polar.

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3 0
3 years ago
How many grams of neutral salt will be obtained in the reaction of calcium oxide with 200 cm 3 of phosphoric acid solution whose
katen-ka-za [31]

Answer:

9.3 g of Ca3(PO4)2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

Next, we shall determine the number of mole of H3PO4 present in 200 cm³ of 0.3 mol/dm³ phosphoric acid (H3PO4) solution. This can be obtained as follow:

Molarity of H3PO4 = 0.3 mol/dm³

Volume = 200 cm³ = 200 cm³/1000 = 0.2 dm³

Mole of H3PO4 =?

Molarity = mole /Volume

0.3 = mole of H3PO4 /0.2

Cross multiply

Mole of H3PO4 = 0.3 × 0.2

Mole of H3PO4 = 0.06 mole

Next, we shall determine the number of mole of the salt, Ca3(PO4)2, obtained from the reaction. This can be obtained as shown below:

3CaO + 2H3PO4 —> Ca3(PO4)2 + 3H2O

From the balanced equation above,

2 moles of H3PO4 reacted to produced 1 mole of Ca3(PO4)2.

Therefore, 0.06 moles of H3PO4 will react to produce = (0.06 × 1)/2 = 0.03 mole of Ca3(PO4)2.

Thus, 0.03 mole of Ca3(PO4)2 is produced from the reaction.

Finally, we shall determine the mass of Ca3(PO4)2 produced as follow:

Mole of Ca3(PO4)2 = 0.03 mole

Molar mass of Ca3(PO4)2 = (40×3) + 2[31 + (16×4)]

= 120 + 2[31 + 64]

= 120 + 2[95]

= 120 + 190

= 310 g/mol

Mass of Ca3(PO4)2 =?

Mole = mass /Molar mass

0.03 = mass of Ca3(PO4)2 / 310

Cross multiply

Mass of Ca3(PO4)2 = 0.03 × 310

Mass of Ca3(PO4)2 = 9.3 g

Thus, 9.3 g of Ca3(PO4)2 was obtained from the reaction.

6 0
2 years ago
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