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Snezhnost [94]
3 years ago
10

How many moles are 2.20 x 10^25 atoms of zinc? Please answer quick will give Brainliest!

Chemistry
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

1.3244 x 10⁴⁹ atoms of zinc

Explanation:

2.20 x 10²⁵ moles

To convert from moles to atoms, we multiply by Avogadro's Number, 6.02 x 10²³

2.20 x 10²⁵ • 6.02 x 10²³

= 1.3244 x 10⁴⁹ atoms of zinc

Hope this helps!

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3. If x electrons are needed to displace 108 g silver from a solution which contains Ag ions,
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Answer:

Choice A. x electrons would be required for displacing 9\; \rm g of aluminum from a solution of \rm Al^{3+} ions.

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Explanation:

The question states that the relative atomic mass of \rm Ag is 108. In other words, each mole of

Therefore, that 108\; \rm g\! of silver that were formed would contain 1\; \rm mol of silver atoms.

Metallic silver would precipitate out of this \rm Ag^{+} solution only after these ions are turned into \rm Ag atoms.

One \rm Ag^{+} ion carries one unit of positive electrical charge. On the other hand, each  e^{-} carries one unit of negative electrical charge.

Therefore, each \rm Ag^{+}\! ion will need to gain one electron to form a neutral \rm Ag atom.

{\rm Ag^{+}}\; (aq) + e^{-} \to {\rm Ag}\; (s).

At least  1\; \rm mol of electrons would be required to turn 1\; \rm mol\! of \rm Ag^{+} ions into that 1\; \rm mol\!\! of silver atoms (which have a mass of 108\; \rm g\!.)

Hence, x = 1\; \rm mol.

Unlike \rm Ag^{+} ions, each aluminum ion \rm Al^{3+} carries three units of positive electrical charge. That is three times the amount of charge on one \rm Ag^{+}\! ion. Therefore, three electrons will be required to turn one \rm Al^{3+}\! ion to an \rm Al atom.

{\rm Al^{3+}}\; (aq) + 3\, e^{-} \to {\rm Al}\; (s)

The question states that the relative atomic mass of \rm Al is 27. Therefore, each mole of \rm Al\! atoms would have a mass 27\; \rm g. There would be \displaystyle \frac{9\; \rm g}{27\; \rm g \cdot mol^{-1}} = \frac{1}{3} \; \rm mol of atoms in that 9\; \rm g of \rm Al\!\!.

It takes 3\; \rm mol of electrons to turn one mole of \rm Al^{3+} ions to one mole of \rm Al atoms. Hence, \displaystyle \frac{1}{3}\times 3\; \rm mol = 1\; \rm mol of electrons would be required to produce that \displaystyle \frac{1}{3}\; \rm mol of \rm Al\! atoms (which has a mass of 9\; \rm g) from \rm Al^{3+}\! ions.

That corresponds to the first choice, x electrons.

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