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n200080 [17]
3 years ago
14

A portion of the periodic table is shown above Which element on the periodic table has properties that are most similar to those

of nitrogen (N) and which element has properties that are the least similar? Explain your reasoning.

Chemistry
2 answers:
fiasKO [112]3 years ago
5 0

Explanation:

Elements that are present in same group contains same number of valence electrons.

Therefore, they tend to show similar chemical properties.

For example, both nitrogen and phosphorus are group 15 elements. Hence, they have 5 valence electrons and they tend to show similar chemical properties.

Since, C, N, P, O and S are all non-metals but Al is a metal.

Therefore, properties of aluminium will be completely different from that of nitrogen or other non-metals.

Hence, we can conclude that element P has properties that are most similar to those of nitrogen and element Al has properties that are the least similar to nitrogen.

liq [111]3 years ago
3 0
Most similar would be phosphorus, because they have the same number of electrons in their outer orbitals, thus they create the same ions, and react with elements in a similar fashion.
Different would probably be aluminum. Aluminum is a metal while nitrogen is a non-metal, as seen by the colors above. Also, aluminum creates a positive ion (losses electrons to become stable) while nitrogen creates a negative ion (gains electrons to become stable)
Hope this helps!!
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3 years ago
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.

Assumption: by "\rm Ag ions" the question meant \rm Ag^{+} with a charge of +1 on each ion.

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.

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What is the percent yield of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? The actua
notka56 [123]

Answer:

(B) 42.1%

Explanation:

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Given: For Na_2S

Given mass = 15.5 g

Molar mass of Na_2S = 78.0452 g/mol

<u>Moles of Na_2S = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>

Given: For CuSO_4

Given mass = 12.1 g

Molar mass of CuSO_4 = 159.609 g/mol

<u>Moles of CuSO_4 = 12.1 g / 159.609 g/mol = 0.0758 moles</u>

According to the given reaction:

Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS

1 mole of Na_2S react with 1 mole of CuSO_4

0.1986 mole of Na_2S react with 0.1986 mole of CuSO_4

Available moles of CuSO_4 = 0.0758 moles

Limiting reagent is the one which is present in small amount. Thus, CuSO_4 is limiting reagent. (0.0758 < 0.1986)

The formation of the product is governed by the limiting reagent. So,

1 mole of CuSO_4 gives 1 mole of CuS

0.0758 mole of CuSO_4 gives 0.0758 mole of CuS

Molar mass of CuS = 95.611 g/mol

Mass of CuS = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g

Theoretical yield = 7.2473 g

Given experimental yield = 3.05 g

% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %

<u>Option B is correct.</u>

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