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Blababa [14]
3 years ago
8

Calculate the pH of each solution at 25∘C

Chemistry
1 answer:
JulijaS [17]3 years ago
8 0

Answer: pH of HCl =5, HNO3 = 1,

NaOH = 9, KOH = 12

Explanation:

pH = -log [H+ ]

1. 1.0 x 10^-5 M HCl

pH = - log (1.0 x 10^-5)

= 5 - log 1 = 5

2. 0.1 M HNO3

pH = - log (1.0 x 10 ^ -1)

pH = 1 - log 1 = 1

3. 1.0 x 10^-5 NaOH

pOH = - log (1.0 x 10^-5)

pOH = 5 - log 1 = 5

pH + pOH = 14

Therefore , pH = 14 - 5 = 9

4. 0.01 M KOH

pOH = - log ( 1.0 x 10^ -2)

= 2 - log 1 = 2

pH + pOH = 14

Therefore, pH = 14 - 2 = 12

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3 years ago
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If 1.320 moles of CH4 reacts completely with oxygen, how many grams of H2O can be formed
Rainbow [258]

Answer:

47.5 g of water can be formed

Explanation:

This is the reaction:

CH₄  +  2O₂  →  CO₂ + 2H₂O

Methane combustion.

In this process 1 mol of methane react with 2 moles of oxygen to produce 2 moles of water and 1 mol of carbon dioxide.

As ratio is 1:2, I will produce the double of moles of water, with the moles of methane I have.

1.320 mol  .2 = 2.64 moles

Now, we can convert the moles to mass (mol . molar mass)

2.64 mol . 18g/mol = 47.5 g

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3 years ago
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During a physical change water can change it's (blank) but it will not change it's (blank).
Valentin [98]
Water can change it's STATE but it will not change it's COMPOUND.
8 0
3 years ago
Determine what is missing from this neutralization reaction: AgNO3+KCl→AgCl+−−−−
lapo4ka [179]

Answer:

The answer to your question is: KNO₃

Explanation:

                                AgNO3  +  KCl   →   AgCl  +   −−−−

A. KNO3    this option is correct because it is a double replacement reaction then potassium must attached to NO₃.

B. KOH  this product is not possible because there is no water to form OH⁻ ions.

C. Ag2K  this product is not possible because both Ag and K are metals, then it is difficult that they attach.

D. KN2O This product is imposible to form, this option is wrong.

7 0
2 years ago
The quantum numbers for the last electron placed in three elements are listed below. Which of these is(are) NOT correct? Er (4 3
UNO [17]

Answer:

The three elements Erbium, Thallium and Osmium have incorrect quantum numbers for the last electron placed.

Explanation:

The 4 quantum numbers are (<em>n,l,ml,ms</em>):

  • <em>n</em> (Principal quantum number): it is the <u>number of the shell (level)</u> where the electron is placed.
  • <em>l </em>(Angular momentum quantum number or Secondary): it represents the <u>sublevel where the electron is</u> placed. There are 4 subleves: s, p d and f so secondary quantum number can take the number 0 (s), 1 (p), 2 (d) or 3 (f) depending on which sublevel the electron is placed.
  • <em>ml</em> (Magnetic quantum number):  it represents the <u>spatial orientation</u> of the electron <u>in respect of the sublevel the electron</u> is placed. For example: if the electron occupies the <em>s sublevel</em> the magnetic number will be <em>0</em>, if the electron occupies the <em>p sublevel</em> the magnetic number could be <em>-1,0,1</em>, if the electron occupies the <em>d sublevel</em> the magnetic number could be <em>-2,-1,0,1,2</em> and if the electron occupies the <em>f sublevel</em> the magnetic number could be <em>-3,-2,-1,0,1,2,3</em>. You can see this in the attachment related to the correct sublevel for the example.
  • <em>ms</em> (Spin quantum number): this number represents the possible rotation of the electron so it could be 1/2 (which is represented by an up arrow) or -1/2 (represented by an down arrow).

Let's analyze the last electron of each element. You can see the attachment for better understanding. The last electron it is represented with orange color.

- Erbium:

This element has 68 electrons so following the Moeller's Diagram to fill the the electronic configuration, we found that the last electron of Erbium it is in the <u>4th level </u>(shell), in the <u>f sublevel</u>. As Erbium has 12 electrons in the f sublevel, it is necessary to follow the Hund's rule (electrons must be placed singly in every sublevel before place a parallel electron) to placed correctly all of them. Finally, the last electron of Erbium stays in the middle of the sublevel and it is represented by a down arrow so the correct quantum numbers in the Erbium element are (4,3,1,-1/2).

- Thallium:

This element has 81 electrons and following the Moeller's Diagram, we found that it last electron it is in the <u>6th level</u>, in the <u>p sublevel</u>. As Thallium has 1 electron in the p sublevel, it is placed singly in the sublevel. So the last electron of Thallium it is represented by an up arrow so the correct quantum numbers in the Thallium element are (6,1,-1,1/2).

- Osmium:

Osmium has 76 electrons and following the steps  that we did with we the other elements, we noticed that its last electron it is in the <u>5th level</u>, in the <u>d sublevel</u>. Following the Hund's rule the last electron of Osmium has a magnetic quantum number of -2 and its spin quantum number is -1/2, so the quantum numbers in the Osmium element are (5,2,-2,-1/2).

<u>Note:</u>

- Remember that the <em>s sublevel</em> has place for 2 electrons, the <u>p sublevel</u> has place for 6 electrons, the <u>d sublevel</u> has place for 10 electrons and the<em> f sublevel</em> has place for 14 electrons.

3 0
3 years ago
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