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gayaneshka [121]
2 years ago
6

One electron is removed from both an Na atom and a K atom, producing two ions. Using principles of atomic structure, explain why

the Na ion is much smaller than the K ion. Discuss both ions in your answer.
Chemistry
2 answers:
olganol [36]2 years ago
7 0

Answer:

Explanation:

The Na ion has 2 electron shells while the K ion  has 3 electron shells making it bigger.

Anni [7]2 years ago
3 0

Answer:

In brief, electrons in the K⁺ ion occupy three main electron shells. In comparison, electrons in the Na⁺ ion occupy only two main electron shells.

Explanation:

Consider the electron cloud model for the structure of an atom or an ion. The nucleus was very dense and small. It was highly located at the center of the atom or the ion. Electrons occupy most of the space of the atom or ion.

If there are more than one electrons, they would be located in one or more main energy levels. Generally, for two atoms/ions with a similar number of protons, the one with more filled main electron shells would have a larger radius.

Electrons in a neutral Na atom occupy three main energy shells. The outermost shell contains only one electron. That electron would be removed first, such that in the Na⁺ ion, electrons would occupy only two main shells.

Similarly, electrons in a neutral K atom occupy four main energy shells. The outermost shell contains only one electron. That electron would be removed first, such that in the K⁺ ion, electrons would occupy three main shells.

Since there are more occupied electrons shells in K⁺ than in Na⁺, K⁺ would have a larger ionic radius.

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A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant? 2Al(s) + 3CuSO4(aq) →
vova2212 [387]

Answer:

Copper (II) sulfate

Explanation:

Given reaction is

2Al(s) + 3CuSO4(aq) → Al2(SO4)3(aq) + 3Cu(s)

Amount of aluminum = 1·25 g

Amount of copper (II) sulfate = 3·28 g

Atomic weight of Al = 26 g

Molecular weight of CuSO4 ≈ 159·5

Number of moles of Al = 1·25 ÷ 26 = 0·048

Number of moles of CuSO4 = 3·28 ÷ 159·5 = 0·021

From the above balanced chemical equation for every 2 moles of aluminum, 3 moles of copper (ll) sulfate will be required

So for 1 mole of Al, 1·5 moles of copper (ll) sulfate will be required

For 0·048 moles of Al, 1.5 × 0·048 moles of copper (ll) sulfate will be required

∴ Number of moles of copper (ll) sulfate required = 0·072

But we have only 0·021 moles of copper (ll) sulfate

As copper (ll) sulfate is not there in required amount, the limiting reactant will be copper (ll) sulfate

∴ The limiting reactant is copper (ll) sulfate

7 0
3 years ago
What is the relationship between the number of gas particles and volume ?
choli [55]

Answer:

V ∝ n  

Step-by-step explanation:

Suppose that pressure and temperature are constant.

If you try to force more molecules of air into a balloon, the balloon will expand.

This is an example of <em>Avogadro's Law</em>: the volume of a gas is directly proportional to the number of moles (particles).

V ∝ n

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A meander is best described as a
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A meander is best described as a bend or curve in a stream channel.
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3 years ago
The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
3 years ago
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