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inessss [21]
4 years ago
15

AMMONIUM CARBONATE

Chemistry
1 answer:
soldier1979 [14.2K]4 years ago
8 0

Answer:

Zero

Explanation:

Hello,

The question require us to calculate the mass of nitrogen present in aluminium carbonate.

This can easily be calculated using Avogadro's number as a constant with some minor calculations but however in this case, we can't because there's no single atom of nitrogen present in aluminium carbonate hence we can't calculate the mass of nitrogen present in it.

Chemical formula of aluminium carbonate = Al₂(CO₃)₃.

From the above chemical formula, we can see that there's no single atom of nitrogen present in the formula hence the mass of nitrogen present in aluminium carbonate that contains 1.23×10²³ carbon atoms is zero.

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Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t
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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}.

It takes three moles of electrons to deposit one mole of gallium atoms \rm Ga\, (s). As a result, \rm 1.89\times 10^{-2}\; mol of electrons would deposit \displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol of gallium atoms \rm Ga\, (s).

8 0
3 years ago
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Answer:

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Explanation:

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