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goldenfox [79]
3 years ago
11

Pick the correct statement about the pure water. Group of answer choices Pure water contains no ions. Pure water contains equal

amounts of hydroxide [OH-] and hydronium [H3O ] ions. Pure water contains larger amounts of hydroxide [OH-] than hydronium [H3O ] ions. Pure water is an electrolyte. Pure water contains smaller amounts of hydroxide [OH-] than hydronium [H3O ] ions.
Chemistry
1 answer:
vekshin13 years ago
5 0

Answer:

Pure water contains no ions.  TRUE

Pure water contains equal amounts of hydroxide [OH-] and hydronium [H3O+ ] ions.  TRUE

Explanation:

This is the equilibrium for pure water:

2H₂O  ⇄  H₃O⁺  +  OH⁻          Kw

We see that pure water has no Ions. Pure water can not conduct electricity.

Generally ionized water comes from the water tap.

Another feature of pure water is pH.

Definetely pH of pure water is : 7

As pH = 7, [H₃O⁺] = 1×10⁻⁷

Then, [OH⁻] = 1×10⁻⁷

This is reazonable because Kw is 1×10⁻¹⁴ and Kw = [H₃O⁺] . [OH⁻]

In conclussion:

Pure water contains no ions.  TRUE

Pure water contains equal amounts of hydroxide [OH-] and hydronium [H3O+ ] ions.  TRUE

Pure water contains larger amounts of hydroxide [OH-] than hydronium [H3O+] ions.  FALSE

Pure water is an electrolyte. FALSE

Pure water contains smaller amounts of hydroxide [OH-] than hydronium [H3O ] ions. FALSE

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The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate
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<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of K_p with K_c is given by the formula:

K_p=K_c(RT)^{\Delta ng}

where,

K_p = equilibrium constant in terms of partial pressure = 3.45

K_c = equilibrium constant in terms of concentration = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature = 500 K

\Delta n_g = change in number of moles of gas particles = n_{products}-n_{reactants}=2-1=1

Putting values in above equation, we get:

3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084

The equation used to calculate concentration of a solution is:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

Initial moles of (CH_3)_3CCl(g) = 1.00 mol

Volume of the flask = 5.00 L

So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

8 0
3 years ago
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