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ryzh [129]
3 years ago
7

When does the energy in the H atom get closer

Chemistry
1 answer:
oee [108]3 years ago
7 0

Answer:

3.48 g

Explanation:

Because its useful.

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Pls help.............
Alina [70]
C is the correct answer
6 0
3 years ago
How many unpaired electrons in ge.
marshall27 [118]

Answer:

Ge only has 2 pair of unpaired electrons.

Explanation:

4 0
3 years ago
Calculate the oxidation number of “C” in CH3COCH3 please
Triss [41]

Answer:

The oxidation number of C (carbon) is +4

Explanation:

8 0
4 years ago
What quantity of copper is deposited by the same quantity of electricity that deposited 9g of aluminum
klio [65]

Answer:

Mass of copper deposited = 31.75 g

Explanation:

According to Faraday's second law of electrolysis, when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of the elements.

From this law, it can be seen that the higher the charge, the lower the number of moles of a given element deposited.

Number of moles of aluminium in 9 g of aluminium = mass / molar mass

Molar mass of aluminium = 27 g

Number of moles of aluminium = 9/27 = 1/3 moles

Charge on aluminium ion = +3

3 moles of electrons will discharge 1 mole of aluminium,

1 mole of electrons will discharge 1/3 moles of aluminium

Number of moles of electrons involved = 1 mole of electrons

Charge on copper ion = +2

1 mole of electrons will discharge 1/2 moles of copper.

Mass of 1/2 moles of copper = number of moles × molar mass of copper

Molar mass of copper = 63.5 g

Mass of copper deposited = 1/2 × 63.5 = 31.75 g

3 0
3 years ago
Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the
Sonbull [250]

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

  • If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If you know pH and pka:

10^(pH-pka) = \frac{[A^-]}{[HA]}

The ratio will be: 10^(pH-pka)

  • At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

0 = log₁₀ \frac{[A^-]}{[HA]}

10^0 = \frac{[A^-]}{[HA]}

1 = \frac{[A^-]}{[HA]}

As ratio is 1,  [conjugate base] = [acid] in solution.

  • At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

  • At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

6 0
3 years ago
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