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VashaNatasha [74]
3 years ago
15

Calculate the OH for a solution that is pH 6.47

Chemistry
1 answer:
katen-ka-za [31]3 years ago
8 0
PH is the negative logarithm of concentration of H+. But there is a relationship between pH and pOH.

pH + pOH = 14
pOH = 14 - pH = 14 - 6.47 = 7.53

Similarly, pOH is the negative logarithm of the concentration of OH- 

pOH = -log(OH-)
7.53 = -log(OH-)
OH- = 2.95 x 10^-6
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Explanation:

Dimensional Analysis is a step by step approach to solving problems in Physics, Chemistry , and Mathematics. It involves having a clear knowledge and understanding to be able to convert a given unit to another in the same dimension using  conversion factors and knowing how they are related to each other.

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Or first approach will be to write out the conversion factor related to our problem which is

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This same process is applied to convert any type of dimensional analysis problems be it physics or mathematics.

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3 years ago
A 4.5-liter sample of a gas has 0.80 mole of the gas. If 0.35 mole of the gas is added, what is the final volume of the gas? Tem
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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

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3 years ago
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3 years ago
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Answer:

0.080 mol

Explanation:

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2.9g*1mol/36.5 g = 0.0795 mol HCl

                              Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

from reaction                            2 mol                   2 mol

given                                       0.0795 mol           x mol

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