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allochka39001 [22]
3 years ago
8

Lithium salts contain a lithium atom that is missing one valence electron (and therefore bears a positive charge). Choose the co

rrect lewis structure of the lithium cation.
Chemistry
2 answers:
olga55 [171]3 years ago
8 0

Answer:

Li⁺

Explanation:

The atomic number of lithium is 3, which means that it has 3 protons and 3 electrons. The electron configuration is [He] 2s¹ and its Lewis structure is Li·

When lithium loses an electron to form lithium cation, its Lewis structue is Li⁺ and it acquires the electron configuration of Helium: 1s².

Roman55 [17]3 years ago
7 0

The atomic number of lithium (Li) is 3.

Thus, its electronic configuration is : 1s^{2}2s^{1}.

Lithium cation is formed when lithium atom loses one of its valence electon.

Hence, [Li+] = 1s^{2}

∴ Lewis dot structure of Li is :

                 Li ·

and lewis dot structure of Lithium cation is:

                        [Li]+


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Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

\text{calculating the initial HI}= \frac{mol}{V}

                                       =\frac{9.3 \times 10 ^ -3}{2}

                                      =0.00465 \ Mol

\text{Similarly}\ \  I_2 \ \  \text{follows} \ \  H_2 = 0 }

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

HI  = 0.00465 - 2x\\\\ I_{2}  \ eq = H_2 \ eq = 0 + x \\\\

It is defined that:

I_2 = 6.29 \times 10^{-4}  \ M \\\\x = I_2 \\\\

HI \ eq= 0.00465 - 2x \\

          =0.00465 -2 \times 6.29 \times 10^{-4} \\\\ =  0.00465 -\frac{25.16 }{10^4}  \\\\   = 0.003392\  M

Now, we calculate the position:  

For the reaction H 2(g) + I 2(g)\rightleftharpoons  2HI(g), you can calculate the value of Kc at 1000 K.  

data expression for Kc

2HI \rightleftharpoons  H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}

         = \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386

calculating the reverse reaction

H_2(g) + I_2(g)\rightleftharpoons  2HI(g)

Kc = \frac{1}{Kc} \\\\

     = \frac{1}{0.034386}\\ \\= 29.081\\

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