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Mnenie [13.5K]
3 years ago
13

When a sodium atom becomes an ion the size of the atom?

Chemistry
1 answer:
jeyben [28]3 years ago
3 0
The size of atom reduces

Electronic configuration of sodium -
2,8,1

In order to become stable it loses one electron and becomes positive ion.
It's size reduces.
You might be interested in
How many miles of C atoms are present in a sample of 3.85x1024 C atoms?
alekssr [168]
Answer:
number of moles = 6.393 moles

Explanation:
One mole of any substance contains Avogadro's number (6.022 * 10^23) of atoms. 
Therefore, to know the number of moles that contain 3.85 * 10^24 atoms, all we have to do is cross multiplication as follows:
1 mole ......................> 6.022 * 10^23
?? moles ..................> 3.85 * 10^24
number of moles = (3.85 * 10^24 *1) / (6.022 * 10^23)
number of moles = 6.393 moles

Hope this helps :)
8 0
2 years ago
Given 25.43 g of O2 gas, how many liters of gas will you have at STP?<br><br>Answer value
Paraphin [41]

Given :

Mass of O₂, m = 25.43 g.

To Find :

How many liters of gas will you have at STP.

Solution :

Molecular mass of O₂, M.M = 2 × 16 gram/mole

M.M = 32 gram/mole

Number of moles is given by :

n = m/M.M

n = 25.43/32 mole

n = 0.79 mole

Hence, this is the required solution.

4 0
3 years ago
Pentanitrogen heptachloride formula
Marianna [84]

Answer:

Explanation:

this is the answer hope it helps!(:

4 0
2 years ago
For the reaction A(g)
zubka84 [21]

Answer:

Yes

Explanation:

By definition, the equilibrium constanct, Kc, for the reaction A ⇒ 2B is

= [A]^1 / [B]^2

Substitute [A] = 4 and [B] = 2 in the equation,

[A]^1 / [B]^2

= 4^1 / 2^2

= 1

= Kc

So yes the reaction is at equilibrium.

6 0
2 years ago
Write the oxidation and reduction half reactions;
luda_lava [24]

Answer:

a)

Fe^{2+}⇒Fe^{3+}+e^-

Br_2+2e^-⇒2Br^-

b)

Mg⇒Mg^{2+}+2e^-

Cr^{3+}+e^-⇒Cr^{3+}

Explanation:

A)

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

Fe^{2+}⇒Fe^{3+}+e^-

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

Br_2+2e^-⇒2Br^-

b)

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

Mg⇒Mg^{2+}+2e^-

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

Cr^{3+}+e^-⇒Cr^{3+}

3 0
3 years ago
Read 2 more answers
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