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Lena [83]
3 years ago
8

A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?

Chemistry
2 answers:
Simora [160]3 years ago
7 0
We determine the concentration of OH- ions by calculating first the pOH first of the solution.

pH + pOH = 14
pOH = 14 - 11.8 = 2.2

pOH = -log (OH-) = 2.2
(OH-) = <span>6.33 x 10^-3 M

Therefore, the correct answer is option B.</span>
balu736 [363]3 years ago
3 0
PH + pOH = 14

11.8 + pOH = 14

pOH = 14 - 11.8

pOH = 2.2

[OH-] = 10 ^- pOH

[OH-] = 10 ^- 2.2

[OH-] = <span>6.33 x 10^-3 M
</span>
Answer B

hope this helps!





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Classify each element. Note that another term for main group is representative, another term for semimetal is metalloid, and the
NikAS [45]

The question is incomplete, here is the complete question:

Classify each element. Note that another term for main group is representative, another term for semi-metal is metalloid, and the inner transition metals are also called the lanthanide and actinide series.

Hf, Am, In, Ta, As, Se, Rn

<u>Answer:</u>

Hafnium and tantalum are transition elements.

Americium is a inner transition element.

Indium, Selenium and Radon are main group elements.

Arsenic is a metalloid.

<u>Explanation:</u>

Main group elements are the elements which belong to s block and p block. They are also known as representative elements.

S-block elements are defined as the elements whose last electron enters s-sub shell. The general electronic configuration of these elements is ns^{1-2}

P-block elements are defined as the elements whose last electron enters p-sub shell. The general electronic configuration of these elements is np^{1-6}

Metalloids are defined as the elements which show intermediate properties between metals and non-metals. There are 7 metalloids in the periodic table. They are: Boron, Silicon, germanium, Arsenic, Antimony, Tellurium and Polonium.

Transition elements are known as d-block elements. D block elements are defined as the elements whose last electron enters d sub shell. The general electronic configuration of these elements is [(n-1)d^{1-10}ns^{0-2}]

Inner transition elements are known as (f block) elements. (F block) elements are defined as the elements whose last electron enters (f subshell). The general electronic configuration of these elements is [(n-2)f^{1-14}(n-1)d^{0-1}ns^{2}]. They are also known as lanthanide and actinide series.

For the given elements:

  • <u>Option 1:</u> Hf

Hafnium is the 72nd element of the periodic table having electronic configuration of [Xe]4f^{14}5d^26s^2

As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 2:</u> Am

Americium is the 95th element of the periodic table having electronic configuration of [Rn]5f^{7}6d^07s^2

As, the last electron is entering the (f subshell), it is a inner transition element.

  • <u>Option 3:</u> In

Indium is the 49th element of the periodic table having electronic configuration of [Kr]5s^25p^1

As, the last electron is entering the p subshell, it is a main group element.

  • <u>Option 4:</u> Ta

Tantalum is the 73rd element of the periodic table having electronic configuration of [Xe]4f^{14}5d^56s^2

As, the last electron is entering the d subshell, it is a transition element.

  • <u>Option 5:</u> As

Arsenic is the 33rd element of the periodic table having electronic configuration of [Ar]4s^24p^3

As, the last electron is entering the p subshell, it is a main group element. It shows an intermediate property of metal and non-metal. Thus, it is a metalloid.

  • <u>Option 6:</u> Se

Selenium is the 34th element of the periodic table having electronic configuration of [Ar]4s^24p^4

As, the last electron is entering the p subshell, it is a main group element.

  • <u>Option 7:</u> Rn

Radon is the 86th element of the periodic table having electronic configuration of [Xe]4f^{14}5d^{10}6s^26p^6

As, the last electron is entering the p subshell, it is a main group element.

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8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?​
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Answer:

the frequency of photons v = 1.509\times10^{39}Hz

Explanation:

Given:  first ionization energy of 1000 kJ/mol.

No. of moles of sulfur = 1 mole

\Delta E_1 = 1000KJ/mol

We know that plank's constant

h = 6.626\times10^{-34} Js

Let the frequency of photons be ν

Also we know that ΔE = hν

this implies ν = ΔE/h

= \frac{10^6J}{6.626\times10^{-34} Js}

v = 1.509\times10^{39}Hz

Hence, the frequency of photons v = 1.509\times10^{39}Hz

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