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Mariulka [41]
3 years ago
12

Which of the following electron configurations is incorrect? A) S [Ne]3s³3p⁴ B) Sn [Kr]5s²4d¹⁰5p² C) Rb [Kr]5s¹ D) V [Ar]4s²3d³

E) I [Kr]5s²4d¹⁰5p⁵
Chemistry
1 answer:
Verizon [17]3 years ago
4 0

Answer:

A S [Ne]3s³3p⁴ instead of [Ne]3s²3p⁴

Explanation:

Hello,

To find the incorrect electronics configuration, we have to refer back to our periodic table or simply writing the electronic configuration of each element down follow principles guiding it such as Aufbau principle and Hund's rule.

a) S = [Ne] 3s³ 3p⁴ instead of [Ne] 3s²3p⁴

Option A is wrong.

S orbital can only accommodate a maximum of 2 electrons and in this case, 3s orbital is carrying 3 electrons. This has violated the rule.

b) Sn = [Kr] 5s² 4d¹⁰ 5p²

Option B is correct

c) Rb = [Kr] 5s¹

Option c is correct

d) V = 4s² 3d³

Option D is correct

e) I = [Kr] 5s² 4d¹⁰ 5p⁵

From the above, we can see that the answer is option A = S

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when Mn2 ions are separated from the mixture, they go through a series of oxidizing and reducing steps. Write the reaction equat
goldfiish [28.3K]

Answer: hello some part of your question is missing below is the missing part

when H₂O and H₂O₂ is added to Mn(OH)₂(s) and put in water bath to dissolve

answer : attached below

Explanation:

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Mn²⁺ ions are separated to the right of the reaction  equations

8 0
3 years ago
Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho
AleksandrR [38]
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
(1/2) ^{n} =x
t_{1/2} = \frac{t}{n}
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>t_{1/2} - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
<span>(1/2) ^{n} =x
</span>(0.5) ^{n} =0.125
n*log(0.5)=log(0.125)
n= \frac{log(0.5)}{log(0.125)}
n=3

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>t_{1/2} = 12.4
</span><span>t_{1/2} *n = t
</span>t= 12.4*3

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
7 0
3 years ago
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Tetraphosphorus Hexoxide
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3 years ago
Complete ionic,net and spectator ions for the following
yawa3891 [41]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

 ZnBr₂ (aq) + AgNO₃ (aq)  →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq)  →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq)  → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq)   →    2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

2) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq)  →   CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq)  + OH₂²⁻  (aq) + 2Na⁺(aq) + SO₄²⁻ (aq)  →   CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq)   + SO₄²⁻ (aq)  →   CaSO₄(s)

The OH⁻ ((aq)  and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

3) Al(NO₃)₃ (aq) + Na₃PO₄ (aq)

Chemical equation:

 Al(NO₃)₃ (aq) + Na₃PO₄ (aq)   → Al(PO₄)(s) + NaNO₃ (aq)

Balanced chemical equation:

Al(NO₃)₃ (aq) + Na₃PO₄ (aq)   → Al(PO₄)(s) + 3NaNO₃ (aq)

Ionic equation:

Al³⁺(aq) + 3NO⁻₃ (aq) + 3Na⁺(aq) + PO₄³⁻ (aq)   → Al(PO₄)(s) + 3Na⁺(aq) + NO⁻₃ (aq)

Net ionic equation:

Al³⁺(aq) + PO₄³⁻ (aq)   → Al(PO₄)(s)

The Na⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The  Al(PO₄) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

4) FeSO₄ (aq) + Ba(OH)₂ (aq)

Chemical equation:

FeSO₄ (aq) + Ba(OH)₂ (aq)  → BaSO₄(s) + Fe(OH)₂(aq)

The equation is already balanced.

Ionic equation:

Fe²⁺(aq)  + SO₄²⁻ (aq) + Ba²⁺(aq)  + 2OH⁻ (aq)  → BaSO₄(s) + Fe²⁺(aq)  + 2OH⁻(aq)

Net ionic equation:

SO₄²⁻ (aq) + Ba²⁺(aq) → BaSO₄(s)

The Fe²⁺ (aq) and OH⁻ (aq) are spectator ions that's why these are not written in net ionic equation. The  BaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

8 0
3 years ago
Where in this image would the Moon need to be in order for there to be a total lunar eclipse?​
sattari [20]

Answer:

A lunar eclipse occurs when the Moon passes directly behind Earth and into its shadow. This can occur only when the Sun, Earth, and Moon are exactly or very closely aligned (in syzygy), with Earth between the other two. A lunar eclipse can occur only on the night of a full moon.

Explanation:

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