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Gala2k [10]
2 years ago
9

Hello people ~

Chemistry
1 answer:
Nonamiya [84]2 years ago
4 0

Answer:

Amphiprotic

Explanation:

  • These chemicals are too independent
  • They can either donate or accept a lone pair of protons .
  • So they can act as base and acid
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Closest noble gas to neon?
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Hi!

I'm not entirely sure about this so I'm sorry if I'm wrong but I think it would be helium.

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What type of tree has thin, waxy needles that drop throughout the year and help to retain water?
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Only coniferous trees have thin, waxy, needle-like leaves that drop throughout the year and help to retain water.

<h3>What are coniferous trees?</h3>

Coniferous trees are evergreen trees that produce cones.

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2 years ago
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t
Sedbober [7]

Answer:

Approximately 6.30\times 10^{-3}\;\rm mol.

Explanation:

The gallium here is likely to be produced from a \rm NaGaO_2\, (aq) solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral "\mathtt{(III)}" next to \rm Ga.  This numeral indicates that the oxidation state of the gallium in this solution is equal to +3. In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of \rm Ga\, (s).

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium \mathtt{(III)} solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s.

Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C.

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}.

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