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Alchen [17]
3 years ago
14

A positively charged ion is called a cation which means there are more protons in the atom compared to the number of protons in

the neutral atom. True False
Chemistry
1 answer:
Lelechka [254]3 years ago
7 0
False False False False False False Ф
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Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
4 years ago
The subshell letter:a) specifies the 3-D shape of the orbital.b) specifies the principal shell of the orbital.c) specifies the p
Dafna11 [192]

Answer:

a) specifies the 3-D shape of the orbital

Explanation:

To determine the region where is more probable to find the electron, it is characterized by four quantum numbers:

- Principal quantum number (n): represents the level, or shell, where the electron is. It varies from 1 to 7, and are represented by the letters K, L, M, N, O, P, and Q.

-Azimuthal quantum number (l): represents the sublevel, or subshell. Is represented by the letters s, p, d, f, g, ... And by the numbers 0, 1, 2, 3, ... Each one has a 3-D shape, the s, for example, has a spherical shape.

-Magnetic quantum number (ml): represents the orbital inside the subshell, and varies from -l to +l, passing by the 0. Each orbital can have at least 2 electrons.

-Spin quantum number (ms): represents the spin of the electron, which can be +1/2 or -1/2.

3 0
3 years ago
What do many bases have in common?
stepladder [879]

Answer:

All bases conduct electricity as they are good electrolytes. All bases turn red litmus paper into blue at the time of indication. Bases have a bitter taste with a soapy texture

Explanation:

hope this helps

7 0
3 years ago
Read 2 more answers
Find the volume of 20g of benzene
sergey [27]

(d) Density of Benzene: 0.8786 g/cm cubed

(m) Mass: 20.00g

Formula to solve (v) volume of benzene: V=m/d

V=20g / 0.8786g/cm cubed

Answer: Volume of benzene is: 22.8 cm cubed  


Explanation: Well, density is mass divided volume. In this, it is what the volume is. All you need is mass divided by density. Hope this helps!

6 0
3 years ago
Organic chemistry involves which type of study?
Mariulka [41]

Answer: option A

Explanation: Organic chemistry deals with the study of carbon and hydrogen containing compounds, drugs and pharmaceuticals etc.

7 0
3 years ago
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