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3 years ago

ASAPP Subtract to find the temperature changes

Chemistry

1 answer:

mamaluj [8]3 years ago

4 0

Hello, I would like to help you, but I really don't understand the question

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

diamong [38]

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

5 0

3 years ago

24.3 2 An artifact classified as seeds, found in a site at Newlands Cross, Ireland, is found to have a 14C radioactivity of 0.10

Rina8888 [55]

Answer:

Age ≅ 7500 years

Explanation:

All radioactive decay is 1st order kinetics and described by the expression

A = A₀e^-kt => t = ln(A/A₀) / -k

k = 0.693 / t(half life) = (0.693 / 5730)yrs⁻¹ = 1.21 x 10⁻⁴ yrs⁻¹

t = Age = [ln(0.103/0.255) / - 1.21 x 10⁻⁴] yrs = 7500 years

5 0

3 years ago

The radius of an atom of gold (Au) is about 1.35 Å.

Eduardwww [97]

Two radius of an atom is equal to the diameter. Adding up all the diameter of the atoms, it should be equal to 9.5 mm. Therefore, we simply convert the units to the same units then divide 1.35 A to 9.5 mm. We calculate as follows:

no. of atoms = 0.0095 m / 1.35x10^-9 m = 7037037 atoms

Hope this answers the question. Have a nice day.

6 0

3 years ago

Read 2 more answers

___Al+NaOH=>__Na3AlO3+H2

MaRussiya [10]

Explanation:

2Al + 2NaOH + 6H2O → 2Na[Al(OH)4] + 3H2

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