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suter [353]
2 years ago
6

How does a lithium cation compare to a lithium atom?

Chemistry
2 answers:
kirill [66]2 years ago
8 0

Lithium cation and Lithium atom can be compared as follow;

<h3>1) Charge:</h3>

              The charge on Lithium atom is zero means, it has same number of protons and electrons. Hence, the nert effect is cancelled out and the atom gets zero charge.

               On the other hand, Lithium cation is formed when it looses one electron. Therefore, the number of protons are greater by one number as compared to number of electrons resulting in the formation of +1 charge as shown below,

                                                Li    →    Li⁺¹  +  e⁻

<h3>2) Size:</h3>

           Secondly, Lithium atom has greater size as compared to Lithium cation because due to presence of greater number of protons compared to electrons in Lithium cation the nuclear charge appears to be greater as compared to neutral atom hence, pulling the valence electrons more effectively making the lithium cation smaller in size as compared to neutral atom.

<h3>3) Reactivity:</h3>

                     We can also compare the reactivity of both species. The neutral will be more reactive than the charged atom because it is containing one valence electron and can donate it when ever it comes in contact with non-metals.

Lerok [7]2 years ago
6 0
Hi Destromceler!

A lithium cation ion is smaller than a lithium atom. In lithium cation it has lost electrons, which causes the remaining electrons to be pulled in stronger by the positive charge in the nucleus. As they get closer to the nucleus decreases the overall size of the atom.

The bigger they are (the more electrons it has) the less effective the proton's pull will be. So if we were talking about a lithium anion (where electrons are gained) then it would be bigger than a lithium atom.


 
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What is the [H+] of a solution with a pH of 9.40?
LiRa [457]

Answer:

a) 3.98 x 10^-10

Explanation:

Hello,

In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:

pH=-log([H^+])

Hence, solving for the concentration of hydronium:

[H^+]=10^{-pH}=10^{-9.40}\\

[H^+]=3.98x10^{-10}M

Therefore, answer is a) 3.98 x 10^-10

Best regards.

4 0
3 years ago
The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this react
Bond [772]

Answer:

The rate of the reaction increased by a factor of 1012.32

Explanation:

Applying Arrhenius equation

ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)

where;

k₂/k₁ is the ratio of the rates which is the factor

Ea is the activation energy = 274 kJ/mol.

T₁ is the initial temperature = 231⁰C = 504 k

T₂ is the final temperature = 293⁰C = 566 k

R is gas constant = 8.314 J/Kmol

Substituting this values into the equation above;

ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)

ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)

ln(k₂/k₁)  = 6.92

k₂/k₁ = exp(6.92)

k₂/k₁ = 1012.32

The rate of the reaction increased by 1012.32

3 0
3 years ago
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A liquid solvent is added to a flask containing an insoluble solid. The total volume of the solid and liquid together is 80.0 mL
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Answer:

111.44 g

Explanation:

We'll begin by calculating the volume of the solvent. This can be obtained as follow:

Mass of solvent = 21 g

Density of solvent = 0.865 g/mL.

Volume of solvent =?

Density = mass /volume

0.865 = 21 /volume of solvent

Cross multiply

0.865 × volume of solvent = 21

Divide both side by 0.865

Volume of solvent = 21 / 0.865

Volume of solvent = 24.28 mL

Next, we shall determine the volume of the solid. This can be obtained as follow:

Volume of solvent + solid = 80.0 mL.

Volume of solvent = 24.28 mL

Volume of solid =?

Volume of solid = (Volume of solvent + solid) – (Volume of solvent)

Volume of solid = 80 – 24.28

Volume of solid = 55.72 mL

Finally, we shall determine the mass of the solid. This can be obtained as follow:

Density of solid = 2.00 g/mL.

Volume of solid = 55.72 mL

Mass of solid =.?

Density = mass / volume

2 = mass of solid / 55.72

Cross multiply

Mass of solid = 2 × 55.72

Mass of solid = 111.44 g

Therefore, the mass of the solid is 111.44 g

8 0
3 years ago
Acid &amp; Base Calculations Calculate the hydronium ion concentration for each. Tell whether it is an acid or a base. 1. pH = 5
Anuta_ua [19.1K]

<u>Explanation:</u>

pH is the negative logarithm of hydronium ion concentration present in a solution.

  • If the solution has high hydrogen ion concentration, then the pH will be low and the solution will be acidic. The pH range of acidic solution is 0 to 6.9
  • If the solution has low hydrogen ion concentration, then the pH will be high and the solution will be basic. The pH range of basic solution is 7.1 to 14
  • The solution having pH equal to 7 is termed as neutral solution.

To calculate the pH of the solution, we use equation:

pH=-\log[H_3O^+]     ......(1)

To calculate the pOH of the solution, we use the equation:

pH + pOH = 14                ........(2)

  • <u>For 1:</u>

We are given:

pH = 5.54

Putting values in equation 1, we get:

5.54=-\log[H_3O^+]

[H_3O^+]=2.88\times 10^{-6}M

Now, putting values in equation 2, we get:

14 = 5.54 + pOH

pOH = 8.46

The solution is acidic in nature.

  • <u>For 2:</u>

We are given:

pOH = 9.7

Putting values in equation 2, we get:

14 = 9.7 + pH

pH = 4.3

Now, putting values in equation 1, we get:

4.3=-\log[H_3O^+]

[H_3O^+]=5.012\times 10^{-5}M

The solution is acidic in nature.

  • <u>For 3:</u>

We are given:

pH = 7.0

Putting values in equation 1, we get:

7.0=-\log[H_3O^+]

[H_3O^+]=1.00\times 10^{-7}M

Now, putting values in equation 2, we get:

14 = 7.0 + pOH

pOH = 7.0

The solution is neither acidic nor basic in nature.

  • <u>For 4:</u>

We are given:

pH = 12.9

Putting values in equation 1, we get:

12.9=-\log[H_3O^+]

[H_3O^+]=1.26\times 10^{-13}M

Now, putting values in equation 2, we get:

14 = 12.9 + pOH

pOH = 1.1

The solution is basic in nature.

  • <u>For 5:</u>

We are given:

pOH = 1.2

Putting values in equation 2, we get:

14 = 1.2 + pH

pH = 12.8

Now, putting values in equation 1, we get:

12.8=-\log[H_3O^+]

[H_3O^+]=1.58\times 10^{-13}M

The solution is basic in nature.

  • <u>For 6:</u>

We are given:

[H_3O^+]=1\times 10^{-5}M

Putting values in equation 1, we get:

pH=-\log(1\times 10^{-5})

pH=5

Now, putting values in equation 2, we get:

14 = 5 + pOH

pOH = 9

The solution is acidic in nature.

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