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3 years ago

Half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution

Chemistry

1 answer:

Elena L [17]3 years ago

5 0

Answer:

2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)

Explanation:

In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.

Reduction reaction occurs on the cathode, in a reduction of water there is gain of 2 electrons to gaseous hydrogen in basic aqueous solution. half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution-

2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)

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You have not mention here about those elements but the general concept for this is the:
uneven distribution ------> polar
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2 years ago

If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass pe

NNADVOKAT [17]

First we need to calculate the number of moles of NaOH titrated.

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number of moles of NaOH = 1.5 × 0.0075 = 0.01125 moles

Then we look at the chemical reaction:

CH $_{3}$ -COOH + NaOH = CH $_{3}$ COONa + H $_{2}$ O

We can see that 1 mole of acetic acid is reacting with one mole of sodium hydroxide. Then we can conclude that 0.01125 moles of sodium hydroxide reacts with 0.01125 moles of acetic acid.

Now we can get the mass of acetic acid:

number of moles = mass (grams) / molecular mass (g/mol)

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mass of acetic acid = 0,01125 × 60 = 0.675 g

We assume that the density of the vinegar = 1 g/mL, so the mass percent of acetic acid is:

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2 years ago

What mass of aluminum has a total nuclear charge of 1.6 CC? Aluminum has atomic number 13. Suppose the aluminum is all of the is

e-lub [12.9K]

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2 years ago

Who has a higher frequency X-ray or, radio waves?

tester [92]

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Explanation:

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2 years ago

A car is traveling at a speed of 65 mi/h. How long does it take to cover a distance of 235 m?

fgiga [73]

Answer:

It will take 8.09 seconds to cover 235 meters by car with speed of 65 mile/h.

Explanation:

Speed of the car = 65 mile/h = 65 1609.34 m/h = 104,607.1 m/h

1 mile = 1609.34 m

Distance covered by the car in t time = 235 m = mile

$Speed-\frac{Distance}{Time}$

$104,607.1 m/h=\frac{\frac{235}{1609.34} mile}{t}$

$t=\frac{235 m}{104,607.1 m/h}$

t = 0.002246 h

1 hour = 3600 seconds

0.002246 h = 0.002246 h × 3600 seconds = 8.09 seconds

It will take 8.09 seconds to cover 235 meters by car with speed of 65 mile/h.

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2 years ago