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Natasha2012 [34]
3 years ago
15

What is the expected oxidation state for the most common ion of element 2

Chemistry
1 answer:
disa [49]3 years ago
8 0
Answer: 1+

Justification:

The ionization energy is the amount of energy needed to loose electrons and becomes ions.

The first ionization energy is the energy needed to liberate one one electron and form the ion with oxidation state 1+.

The second ionization energy is the energy to release a second electron and form the ion with oxidation state 2+.

The third ionization energy is the energy to leave a third electron free and form the ion with oxidation state 3+.

The relatively low first ionization energy of element 2, shows it  it will lose an electron easily to form the ion with oxidations state 1+.

The second and third ionization energies are very high meaning that the ions with oxidation staes 2+ and 3+ will not be formed.

Therefore, the answer is that the expected oxidation state for the most common ion of element 2 is 1+.
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A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
4 years ago
Read 2 more answers
If one-trillionth of the atoms of a radioactive isotope disintegrate each day, what is the decay constant of the process?
Paul [167]

Decay constant of the process 1×10^(-12) day^(-1).

<h3>What is decay constant?</h3>

A radioactive nuclide's probability of decay per unit time is known as its decay constant, which is expressed in units of s1 or a1. As a result, as shown by the equation dP/P dt =, the number of parent nuclides P declines with time t. Nuclear forces are about 1,000,000 times more powerful than electrical and molecular forces in their ability to bind protons and neutrons. The strength of the bonds holding the radioactive element are likewise indifferent to the decay probabilities and's, in addition to being unaffected by temperature and pressure. The decay constant is related to the nuclide's T 1/2 half-life by T 1/2 = ln 2/.

To know more about decay constant:

brainly.com/question/16623902

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3 0
2 years ago
How much of each gas is there in 100cm3 air?
Akimi4 [234]
Each 100 cm3 of air, constitutes 78cm3 nitrogen, 21cm3 oxygen and 1cm3 constitutes of other gases like Argon, ozone, carbon dioxide and water vapour in small amounts.
8 0
4 years ago
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For the following H30 concentration list the pH and the OH concentration in
Alenkinab [10]

Answer:

[H3O+] = 1.4*10^-5 M

pH = 4.85

[OH-] = 7.08*10^-10

Explanation:

As pH is a measure of hydronium H3O concentration, simply substitute [H3O+] into the following equation:
pH = -log[H+]
pH = -log(1.4*10^-5)
pH = 4.853871...
Round to 2-3 sig figs due to only being given data with 2 significant figures
pH = 4.85

One method to get to [OH-] is to turn pH into pOH and then use inverse functions to get [OH-]
pH + pOH = 14
4.85 + pOH = 14
pOH = 9.15

Then to get [OH-] from pOH:
pOH = -log[OH-]
9.15 = -log[OH-]
-9.15 = log[OH-]
10^(-9.15) = [OH-]
7.07945784 * 10^-10 = [OH]
Round based on given significant figures again:
7.08*10^-10 = [OH-]

(Feel free to add any questions & I'll be sure to reply if clarification is needed)

6 0
2 years ago
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Fill in the Blanks :
MariettaO [177]

Answer:

Explanation:

1. Mass number

2. Protons

3. Electrons

8 0
3 years ago
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