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lilavasa [31]
3 years ago
7

Match each transition metal ion with its condensed ground-state electron configuration. A [Ar]3d2 B [Ar]4s23d3 C [Kr]4d10 D [Xe]

E [Ar]3d6 F [Ar]3d5 G [Xe]4f145d10 Mn2+ Hg2+ La3+ Fe3+ Ag+ Co3+
Chemistry
1 answer:
madreJ [45]3 years ago
7 0

Answer:

Mn^{2+} : - F . [Ar]3d^{5}

Hg^{2+} : - G. [Xe]4f^{14}5d^{10}

La^{3+} : - D. [Xe]

Fe^{3+} : - F. [Ar]3d^{5}

Ag^{+} : - C. [Kr]4d^{10}

Co^{3+} : - E. [Ar]3d^{6}

Explanation:

The electronic configuration of the element Mn is:-

[Ar]3d^{5}4s^2

For, Mn^{2+}, 2 electrons are lost, thus the configuration is:-

[Ar]3d^{5}

The electronic configuration of the element Hg is:-

[Xe]4f^{14}5d^{10}6s^2

For, Hg^{2+}, 2 electrons are lost, thus the configuration is:-

[Xe]4f^{14}5d^{10}

The electronic configuration of the element La is:-

[Xe]5d^{1}6s^2

For, La^{3+}, 3 electrons are lost, thus the configuration is:-

[Xe]

The electronic configuration of the element Fe is:-

[Ar]3d^{6}4s^2

For, Fe^{3+}, 3 electrons are lost, thus the configuration is:-

[Ar]3d^{5}

The electronic configuration of the element Ag is:-

[Kr]4d^{10}5s^1

For, Ag^{+}, 1 electron is lost, thus the configuration is:-

[Kr]4d^{10}

The electronic configuration of the element Co is:-

[Ar]3d^{7}4s^2

For, Co^{3+}, 3 electrons are lost, thus the configuration is:-

[Ar]3d^{6}

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Explanation:

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8 0
3 years ago
Empirical formula of a compound composed of 32.1 g potassium (k) and 6.57 g oxygen (o)?
tester [92]

The empirical formula is K₂O.

The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.

The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.

So, our job is to calculate the <em>molar ratio</em> of K to O.

Step 1. Calculate the <em>moles of each element </em>

Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K

Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0

Step 2. Calculate the <em>molar ratio of each elemen</em>t

Divide each number by the smallest number of moles and round off to an integer

K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1

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5 0
3 years ago
Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
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White raven [17]

Answer:

False

Explanation:

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