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Inessa05 [86]
3 years ago
7

What element’s valence electrons are located in 5s25p4?

Chemistry
1 answer:
IrinaK [193]3 years ago
5 0
The element is TELLURIUM. Tellurium has atomic number 52 with the electronic configuration of [Kr] 4d10 5s2 5p4. The element belongs to group 16 elements on the periodic table. It exhibits various oxidation states: +2, +4, +6, and -2. +4 is its most common oxidation state.
You might be interested in
Gallium is produced by the electrolysis of a solution made by dissolving gallium oxide in concentrated NaOH ( aq ) . Calculate t
Sedbober [7]

Answer:

Approximately 6.30\times 10^{-3}\;\rm mol.

Explanation:

The gallium here is likely to be produced from a \rm NaGaO_2\, (aq) solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?

Note the Roman Numeral "\mathtt{(III)}" next to \rm Ga.  This numeral indicates that the oxidation state of the gallium in this solution is equal to +3. In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of \rm Ga\, (s).

As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium \mathtt{(III)} solution.

How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.

t = \rm 80.0\; min = 80.0\; min \times 60\;s \cdot min^{-1} = 4800\; s.

Q = I \cdot t = \rm 0.380 \; A \times 4800 \; s = 1.824\times 10^3\; C.

Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.

\begin{aligned} n(\text{electrons}) &= \frac{Q}{F} \cr &= \rm \dfrac{1.824\times 10^3\; C}{96485.332\; C \cdot mol^{-1}}\cr &\approx \rm 1.89\times 10^{-2}\; mol \end{aligned}.

It takes three moles of electrons to deposit one mole of gallium atoms \rm Ga\, (s). As a result, \rm 1.89\times 10^{-2}\; mol of electrons would deposit \displaystyle \rm \frac{1}{3}\times 1.89\times 10^{-2}\; mol \approx 6.30\times 10^{-3}\; mol of gallium atoms \rm Ga\, (s).

8 0
3 years ago
Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid (HC3H5O2, Ka= 1.3x10-5 ) (b) 0.1000M sodium prop
jek_recluse [69]

(a) The pH of 0.1000 M propanoic acid (HC3H5O2) is 2.9.

(b) The pH of 0.1000 M sodium propanoate (NaC3H5O2) is 8.9.

(c) The pH of 0.1000 M propanoic acid (HC3H5O2) and 0.1000 M sodium propanoate (NaC3H5O2) is 4.9.

<h3>Further explanation:</h3>

(a)

Given information:

The value of acid ionization constant for propanoic acid is  1.3 x 10^{-5} .

The initial concentration of propanoic acid is  .

To calculate:

The pH of 0.1000 M propanoic acid solution.

Solution:

Propanoic acid  is a weak acid. It ionizes partially in water as follows:

 

The expression for acid dissociation constant is,

                                                            …… (1)

Here,

 is ionization constant of propanoic acid.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydronium ion.

 is the equilibrium concentration of propanoic acid.

ICE table (1):

 

Refer ICE table (1),

 

Substitute the values form the ICE table (1) in equation (1).

 

The approximation x is very small is valid. Therefore, the value of x can be neglected. Above equation can be modified as,

 

Rearrange above equation for x.

                                                                                                           …… (2)

Substitute   for   in equation (2) to calculate the value of x.

 

Therefore, from the ICE table (1) the concentration of hydronium ion is,

 .

The negative logarithm of hydronium ion concentration is defined as the pH of the solution. Mathematically,

                                                                                                               …… (3)

Substitute    for    in equation (3) to calculate the pH of the solution.

 

(b)

Given information:

The value of acid ionization constant for propanoic acid is  .

The initial concentration of sodium propanoate is  .

To calculate:

The pH of 0.1000 M sodium propanoate solution.

Solution:

Sodium propanoate  is conjugate base of weak propanoic acid. It undergoes hydrolysis in water to yield hydroxide ion in the solution as follows:

                                                        …… (4)

Propanoic acid  is a weak acid. It ionizes partially in water as follows:

                                                       …… (5)

Dissociation reaction for water is written as follows:

                                                                                       …… (6)

From equation (4), (5), and (6) the relationship between   and   is,

                                                                                                                              …… (7)

Substitute   for   and   for   in equation (7).

 

ICE table (2):

 

The expression for base dissociation constant is,

                                                                                                     …… (8)

Here,

is base ionization constant.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydroxide ion.

 is the equilibrium concentration of propanoic acid.

From the ICE table (2),

 

Substitute the values form the ICE table (2) in equation (8).

 

The approximation y is very small is valid. Therefore, the value of y can be neglected. Above equation can be modified as,

 

Rearrange above equation for y.

                                                                                                           …… (9)

Substitute   for   in equation (9) to calculate the value of y.

 

Therefore, from the ICE table (2) the concentration of hydroxide ion is,

 

The negative logarithm of hydroxide ion concentration is defined as pOH of the solution. Mathematically,

                                                                                                           …… (10)

Substitute    for    in equation (10) to calculate pOH of the solution.

 

The relation between pH and pOH is as follows:

                                                                                                                   …… (11)

Substitute 5.057 for pOH in equation (11) to calculate the pH of the solution.

 

(c)

Given information:

The value of acid ionization constant for propanoic acid is  .

The initial concentration of sodium propanoate is  .

The initial concentration of sodium propanoate is  .

To calculate:

The pH of 0.1000 M sodium propanoate and 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid, and sodium propanoate is salt of the conjugate base of propanoic acid. Thus, propanoic acid and sodium propanoate will form a buffer system.

The pH of the buffer solution can be determined with the help of the Henderson-Hasselbalch equation. Mathematically,

 

For propanoic acid and sodium propanoate buffer system, the Henderson-Hasselbalch equation can be modified as,

                                                                                               …… (12)

The negative logarithm of acid ionization constant is equal to  .

                                                                                                                …… (13)

Substitute   for  in equation (13).

 

Substitute    for  ,   for   and 4.9 for    in equation (12).

 

Learn more:

1. About Henderson-Hasselbalch equation brainly.com/question/12999557

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ionic equilibria

Keywords: ionic equilibrium, propanoic acid, sodium propanoate, ionization constant, weak acid, conjugate base, equilibrium concentration, hydronium ion, hydroxide ion, pH, pOH, ICE table, negative logarithm, buffer solution, Henderson-Hasselbalch equation, 0.1000 M, 4.9, 8.9, 2.9.

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3 years ago
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Why are group 1a (1) and group 2a (2) elements found in many compounds, but not group 8a (18) elements?
Rashid [163]
Because group 1a and 2a are most reactive due to their least number of electrons in their outmost shell compared to group 8a(18)
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14. A cinder cone is the simplest type of volcano.<br> a. True<br> b. False
dusya [7]

Answer:

True

Explanation:

They are built from particles and blobs of congealed lava ejected from a single vent. As the gas-charged lava is blown violently into the air, it breaks into small fragments that solidify and fall as cinders around the vent to form a circular or oval cone, making them the simplest out there

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What is the chemical formula of calcium chloride?
Luba_88 [7]
<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ It would be CaCl2

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

7 0
3 years ago
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