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Leya [2.2K]
3 years ago
12

Calculate the pH of lemon juice to three significant figures, with a hydrogen ion concentration of

Chemistry
1 answer:
Airida [17]3 years ago
3 0

Answer:

pH = - log [2.12 x 10^-3]

what is the log of 2.22 x 10^-3]

Take the opposite of that,

That is the pH, now, just make certain you use the correct significant figures.

Explanation:

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Removing the tip of the stem in a young plant will most directly interfere with the production of
denis23 [38]

Answer:

c.

Explanation:

4 0
3 years ago
When a aqueous solution of a certain acid is prepared, the acid is dissociated. Calculate the acid dissociation constant of the
stepan [7]
<h2>K_a = \dfrac{[H^{+}] [A^{-}]}{[HA]}</h2>

Explanation:

  • When an aqueous solution of a certain acid is prepared it is dissociated is as follows-

        {\displaystyle {\ce {HA  ⇄  {H^+}+{A^{-}}}  }}

Here HA is a protonic acid such as acetic acid, CH_3COOH

  • The double arrow signifies that it is an equilibrium process, which means the dissociation and recombination of the acid occur simultaneously.
  • The acid dissociation constant can be given by -

        K_a = \dfrac{[H^{+}] [A^{-}]}{[HA]}

  • The reaction is can also be represented by Bronsted and lowry -

         \\{\displaystyle {\ce {{HA}+ H_2O} ⇄  [H_3O^+] [A^-]

  • Then the dissociation constant will be

        K_a = \dfrac{[H_3O^{+}] [A^{-}]}{[HA]}

Here, K_a is the dissociation constant of an acid.

6 0
3 years ago
How many bonding pairs in br2o
Evgen [1.6K]
There are 22 bonding parts
5 0
3 years ago
opper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f =
Y_Kistochka [10]

Answer:

55.373g/l

Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

The solubility product of CuBr is 6.3×10−9.

The concentration of NH3 is 0.10 M.

Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

Suppose the solubility of CuBr is “s”.

It is given that concentration of NH3 is 0.10 M.

The equilibrium constant expression for the above reaction is as follows,

Here,

The concentration of pure solids is 1 M. Thus, the concentration of CuBr is 1 M.

As calculated, the value of Ksp is 396.9.

Substitute all the required values in above formula.  

On further solving above equation,

Therefore, the solubility of CuBr in ammonia is 0.386 M.

The formula to calculate solubility

Solubuility (g/l)= Molarity(M) x Molarmass

Chemistry homework question answer, step 2, image 10

The molar mass of CuBr is 143.45 g/mol.

The formula to calculate solubility in g/L is given below.

The molar mass of CuBr is 143.45 g/mol.

therefore,

solubility = 0.386M x 143.45g/mol

where (M = mol/l)

solubility = 55.375g/l

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3 years ago
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