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aleksley [76]
3 years ago
7

Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in

Chemistry
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

Redox

Explanation:

Reduction is gain of electrons

oxidation is loss of electrons

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The Great Plains are also referred to as____.
Gnesinka [82]

Answer:

The interior plains

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2 years ago
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Your mother has an unlabeled bottle of clear liquid in the refrigerator. You pour a small cup of the liquid and taste it, and it
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The pH of the liquid would be 6 or a 5
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What will the pressure of H+ ions in a solution cause
suter [353]

ACIDIC BEHAVIOR OF SOLUTION

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3 years ago
Hydrogen produced from a hydrolysis reaction was collected over water and the following data was compiled.
Shkiper50 [21]

Answer:

  • 0.00358 mol

Explanation:

<u>1) Data:</u>

a) V = 93.90 ml

b) T = 28°C

c) P₁ = 744 mmHg

d) P₂ = 28.25 mmHg

d) n = ?

<u>2) Conversion of units</u>

a) V = 93.90 ml × 1.000 liter / 1,000 ml = 0.09390 liter

b) T = 28°C = 28 + 273.15 K = 301.15 K

c) P₁ = 744 mmHg × 1 atm / 760 mmHg = 0.9789 atm

d) P₂ = 28.5 mmHg × 1 atm / 760 mmHg = 0.0375 atm

<u>3) Chemical principles and formulae</u>

a) The total pressure of a mixture of gases is equal to the sum of the partial pressures of each gas. Hence, the partical pressure of the hydrogen gas collected is equal to the total pressure less the vapor pressure of water.

b) Ideal gas equation: pV = nRT

<u>4) Solution:</u>

a) Partial pressure of hydrogen gas: 0.9789 atm - 0.0375 atm = 0.9414 atm

b) Moles of hygrogen gas:

pV = nRT ⇒ n = pV / (RT) =

n =  (0.9414 atm × 0.09390 liter) / (0.0821 atm-liter /K-mol × 301.15K) =

n = 0.00358 mol (which is rounded to 3 significant figures) ← answer

7 0
3 years ago
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A geochemist in the field takes a 46.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.
inysia [295]

Answer:

The solubility of the mineral compound X in the water sample is 0.0189 g/mL.

Explanation:

Step 1: Given data

The volume of water sample = 46.0 mL.

The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.

Step 2: Calculate the solubility of X in water

46.00 mL of water sample contains 0.87 g of the mineral compound X.

To calulate how many grams of the mineral compound  1.0 mL  of water sample contains:

0.87 g/46.0 mL = 0.0189 g.

This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.

3 0
3 years ago
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