All categories

3 years ago

The symbol for xenon xe would be part of the noble gas notation for the element

2 answers:

8 0

The symbol for xenon xe would be part of the noble gas notation for the element of radium, it is a chemical element with the symbol of Ra and the atomic number of radium is 88. It is the sixth element in group 2 of the periodic table, also known as the alkaline earth metals.<span> </span>

Mademuasel [1]3 years ago

4 0

Answer : The noble gas notation is often given to the short or abbreviated form of the electron configuration.

The use of symbol of the previous noble gas as part of the electron configuration of an element is the shortcut way to write the electron configuration.

Xenon has electron configuration as $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6}$ .

Cesium has the electron configuration as $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 6s^{1}$

So, Cesium can be abbreviated using the symbol of xenon [Xe] $6s^{1}$ .

You might be interested in

Which balance is a centigram balance? <br> A <br> Or <br> B

MaRussiya [10]

I think it is B because you didn’t type the options

5 0

3 years ago

Complete the following precipitation reactions with balanced molecular, total ionic, and net ionic equations:

Dmitry [639]

Molecular equation

Hg₂(NO₃)₂ (aq) + KI(aq) ⇒Hg₂I₂(s) + 2KNO₃(aq)

Total Ionic equation

Hg²⁺(aq) + 2NO³⁻(aq) + 2K⁺aq) ⇒Hg₂I₂(s) + 2K⁺(aq) + NO³⁻ (aq)

Net Ionic equation

Hg²⁺(aq) + 2I⁻(aq) ⇒ Hg₂I₂(s)

<h3>What is the molecular equation?</h3>

Sometimes, a balanced equation is all that is used to refer to a chemical equation. Any ionic substances or acids are represented using their chemical formulas as neutral compounds in a molecular equation. Each substance's state is described in parenthesis after the formula. A complete ionic equation also contains the spectator ions, whereas a net ionic equation just displays the chemical species that are involved in a reaction.

The steps listed below can be used to determine the net ionic equation for a specific reaction:

Include the states of each chemical in the balanced molecular equation for the reaction.

To know more about the molecular equation, visit:

brainly.com/question/14286552

#SPJ4

4 0

1 year ago

If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?

aksik [14]

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

Mn(OH)₂ → Mn⁺² + 2OH⁻

[Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

Ksp = [Mn⁺²][OH-]²

[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

[OH⁻]² = 10⁻³

[OH⁻] = (10⁻³)¹⁽²

[OH⁻] = 0.0316 M

Calculate the pOH

pOH = -log [OH⁻]

pOH = -log [0.0316]

pOH = 1.5

Now calculate pH

pH = 14 - pOH

pH = 14 - 1.5

pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

#SPJ4

7 0

2 years ago

Would you see more living things in a Biome or an Ecosystem?

kicyunya [14]

Answer:

A Biome is way bigger than a ecosystem

Explanation:

5 0

3 years ago

Read 2 more answers

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first i

Law Incorporation [45]

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹; 5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹: s s

$K_{sp} =\text{[Ag$^{+}$][IO$_{3}$$^{-}$]} = s\times s = s^{2} = 3.0\times 10^{-8}\\s = \sqrt{3.0\times 10^{-8}} \text{ mol/L} = 1.7 \times 10^{-4} \text{ mol/L}$

b. Barium sulfate

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹: 0.02 0

C/mol·L⁻¹: +s +s

E/mol·L⁻¹: 0.02 + s s

$K_{sp} =\text{[Ba$^{2+}$][SO$_{4}$$^{2-}$]} = (0.02 + s) \times s \approx 0.02s = 1.1\times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{0.02} \text{ mol/L} = 5.5 \times 10^{-9} \text{ mol/L}$

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is

$\mu = \dfrac{1}{2} \sum_{i} {c_{i}z_{i}^{2}}\\\\\mu = \dfrac{1}{2} (\text{[Ba$^{2+}$]}\cdot (2+)^{2} + \text{[NO$_{3}$$^{-}$]}\times(-1)^{2}) = \dfrac{1}{2} (\text{0.02}\times 4 + \text{0.04}\times1)= \dfrac{1}{2} (0.08 + 0.04)\\\\= \dfrac{1}{2} \times0.12 = 0.06$

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(1)^{2}\sqrt{0.06} = -0.51\times 0.24 = -0.12\\\gamma = 10^{-0.12} = 0.75$

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

$K_{sp} =\text{[Ag$^{+}$]$\gamma_{Ag^{+}}$[IO$_{3}$$^{-}$]$\gamma_{IO_{3}^{-}}$} = s\times0.75\times s \times 0.75 =0.56s^{2}= 3.0 \times 10^{-8}\\s^{2} = \dfrac{3.0 \times 10^{-8}}{0.56} = 5.3 \times 10^{-8}\\\\s =2.3 \times 10^{-4}\text{ mol/L}$

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

$\log \gamma = -0.51z^{2}\sqrt{I} = -0.051(2)^{2}\sqrt{0.06} = -0.51\times16\times 0.24 = -0.50\\\gamma = 10^{-0.50} = 0.32$

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

$K_{sp} =\text{[Ba$^{2+}$]$\gamma_{ Ba^{2+}}$[SO$_{4}$$^{2-}$]$\gamma_{ SO_{4}^{2-}}$} = (0.02 + s) \times 0.32\times s\times 0.32 \approx 0.02\times0.10s\\2.0\times 10^{-3}s = 1.1 \times 10^{-10}\\s = \dfrac{1.1\times 10^{-10}}{2.0 \times 10^{-3}} \text{ mol/L} = 5.5 \times 10^{-8} \text{ mol/L}$

7 0

3 years ago