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Elenna [48]
3 years ago
12

write equations to show the chemical processes which occur when the first ionization and the second ionization energies of lithi

um are measured? ​
Chemistry
1 answer:
diamong [38]3 years ago
5 0

Answer:

First ionization of lithium:

\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}.

Second ionization of lithium:

\text{Li}^{+}\;(g) \to\text{Li}^{2+} \;(g) + \text{e}^{-}.

Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

  • Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol (g) next to any atom or ion in the equation.
  • The product shall contain one gaseous ion and one electron. The charge on the ion shall be the same as the order of the ionization energy. For the second ionization energy, the ion shall carry a charge of +2.
  • Charge shall balance on the two sides of the equation.

First Ionization Energy of Li:

  • The products shall contain a gaseous ion with charge +1 \text{Li}^{+}\;(g) as well as an electron \text{e}^{-}.
  • Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: \text{Li}\;(g).
  • Hence the equation: \text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}.

Second Ionization Energy of Li:

  • The product shall contain a gaseous ion with charge +2: \text{Li}^{2+}\;(g) as well as an electron \text{e}^{-}.
  • Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?
  • The equation for this process is \text{Li}^{+} \; (g) \to \text{Li}^{2+}\;(g) + \text{e}^{-}.
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250 mL of a solution of calcium oxalate is the evaporated until only a residue of solid calcium
Lana71 [14]

Answer:

2.3 * 10^-5

Explanation:

Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.

Hence;

Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols

From the question;

1.2 * 10^-3 mols dissolves in 250 mL

x moles dissolves in 1000mL

x = 1.2 * 10^-3 mols * 1000/250

x= 4.8 * 10^-3 moldm^-3

CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)

Hence Ksp = [Ca^2+] [C2O4^2-]

Where;

[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3

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2 years ago
At which type of tectonic plate boundary is a volcano least likely to occur?
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The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc
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Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles

b) moles of C_2H_4

\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles

H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)

According to stoichiometry :

1 mole of C_2H_4 combine with 1 mole of H_2

Thus 0.33 mole of C_2H_4 will combine with =\frac{1}{1}\times 0.33=0.33 mole of H_2

Thus C_2H_4 is the limiting reagent as it limits the formation of product.

As 1 mole of C_2H_4 give =  1 mole of C_2H_6

Thus 0.33 moles of C_2H_4 give =\frac{1}{1}\times 0.33=0.33moles  of C_2H_6

Mass of C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g

Thus theoretical yield (g) of C_2H_6 produced by the reaction is 9.9 grams

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