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Rudik [331]
3 years ago
11

Which statement is TRUE about oxidation-reduction reactions? a. Every oxidation must be accompanied by a reduction. b. There are

four commonly accessed oxidation states of carbon. c. Dehydrogenases typically remove two electrons and two hydrides. d. They usually proceed through homolytic cleavage. e. During oxidation a compound gains electrons.
Chemistry
1 answer:
Tomtit [17]3 years ago
8 0

Answer:

Every oxidation must be accompanied by a reduction.

Explanation:

Oxidation and reduction are complementary processes. There can be no oxidation without reduction and vice versa. It is actually a given an take affair. A specie looses electrons which must be gained by another specie to complete the process. This explains why the selected option is the correct one.

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When copper metal is added to nitric acid, the following reaction takes place
zlopas [31]

Answer:

The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.

Explanation:

Cu (s) + 4 HNO_3 (aq) \rightarrow Cu(NO_3)_2 (aq) + 2 H_2O (l) + 2 NO_2 (g)

Moles of copper = \frac{2.01 g}{63.55 g/mol}=0.03163 mol

According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.

Then 0.03613 moles of copper will give:

\frac{2}{1}\times 0.03163 mol=0.06326 mol of nitrogen dioxide gas

Moles of nitrogen dioxide gas = n = 0.06326 mol

Pressure of the gas = P

P = Total pressure - vapor pressure of water

P = 726 mmHg - 23.8 mmHg = 702.2 mmHg

P = 0.924 atm (1 atm = 760 mmHg)

Temperature of the gas = T = 25.0°C =298.15 K

Volume of the gas = V

PV=nRT

V=\frac{0.06326 mol\times 0.0821 atm L/mol K\times 298.15 K}{0.924 atm}

V = 1.68 L

The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.

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2 years ago
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how many grams of phosphorus (P4) react with 35.5L of O2 at STP to form solid tetraphosphorus decaoxide
algol [13]

Answer:

mass P4 = 35.998 g

Explanation:

  • P4 + 5O2 → P4O10

∴ STP: P = 1 atm; T = 298 K

∴ V O2= 35.5 L

⇒ nO2 = P.V / R.T

∴ R = 0.082 atm.L/K.mol

⇒ nO2 = ((1 atm)×(35.5L))/((0.082 atm.L/K.mol)(298K))

⇒ nO2 = 1.453 mol O2

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