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3 years ago

1) What is the ground state electron configuration for Oxygen (O)?

Chemistry

1 answer:

statuscvo [17]3 years ago

6 0

Answer:

1)Oxygen atoms have 8 electrons and the shell structure is 2.6. The ground state electron configuration of ground state gaseous neutral oxygen is [He]. 2s2. 2p4 and the term symbol is 3P2.

Explanation:

2)If the element were to become excited, the electron could occupy an infinite number of orbitals. However, in most texts the example will be the next available one. So for oxygen, it might look like this: 1s22s22p33s1 - where the valence electron now occupies the 3s orbital in an excited (i.e. not ground) state

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Be sure to answer all parts.The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orange-yellow compound, fro

miss Akunina [59]

Explanation:

Since, the given reaction is as follows.

$2NO(g) + Cl_{2} \rightleftharpoons 2NOCl$

Expression for $Q_{c}$ of this reaction is as follows.

$Q_{c} = \frac{[NOCl]^{2}}{[NO]^{2}[Cl_{2}]}$

Putting the given values into the above formula as follows.

$Q_{c} = \frac{[NOCl]^{2}}{[NO]^{2}[Cl_{2}]}$

= $\frac{(9.50)^{2}}{(4.40 \times 10^{-2})^{2} \times (1.80 \times 10^{-3})}$

= $\frac{90.25}{34.848 \times 10^{-7}}$

= $2.59 \times 10^{-7}$

Thus, we can conclude that $Q_{c}$ for the experiment is $2.59 \times 10^{-7}$ .

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4 years ago

How to prove that a given colorless liquid is an acid

Pachacha [2.7K]

Answer:

Explanation:

by using litmus papaer it will turn into red if solutions is acidic

and into blue if it's basic

by using Magnesium ribbon. Magnesium do not react with bases so hydrogen gas will produce when Magnesium is dipped in an acidic solution

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3 years ago

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Olin [163]

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3 years ago

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A pressure of 125,400 pa is equal to what kPa

Goryan [66]

Answer:

125.4 kilopascals

Explanation:

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3 years ago

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PH is a logarithmic scale used to indicate the hydrogen ion concentration, [H+], of a solution: pH=−log[H+] Due to the autoioniz

ruslelena [56]

Answer:

A) pH = 2.42
B) pH = 12.00

Explanation:

The dissolution of HCl is HCl → H⁺ + Cl⁻

To solve part A) we need to calculate the concentration of H⁺, to do that we need the moles of H⁺ and the volume.

The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:

$0.35g_{HCl}*\frac{1mol_{HCl}}{36.46g_{HCl}} *\frac{1molH^{+}}{1mol HCl}$ = 9.60*10⁻³ mol H⁺

So the concentration of H⁺ is

[H⁺] = 9.60*10⁻³ mol / 2.5 L = 3.84 * 10⁻³ M

pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42

The dissolution of NaOH is NaOH → Na⁺ + OH⁻
Now we calculate [OH⁻], we already know that V = 2.0 L, and a similar process is used to calculate the moles of OH⁻, keeping in mind the molecular weight of NaOH, 40 g/mol:

$0.80g_{NaOH}*\frac{1mol_{NaOH}}{40g_{NaOH}} *\frac{1molOH^{-}}{1mol NaOH}$ = 0.02 mol OH⁻

[OH⁻] = 0.02 mol / 2.0 L = 0.01

pOH = -log [OH⁻] = -log (0.01) = 2.00

With the pOH, we can calculate the pH:

pH + pOH = 14.00

pH + 2.00 = 14.00

pH = 12.00

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3 years ago