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3 years ago

If a solution is strongly acidic, what anionscannot be present in appreciableconcentrations?

Chemistry

1 answer:

masha68 [24]3 years ago

7 0

Explanation:

$OH^-$ cannot be present in appreciable concentrations in a solution that is strongly acidic in nature. A weak acid is a complex that consists of hydrogen bound to an anion that does not dissociate well in solution comparing to a strong acid which dissolves completely to an anion. Therefore, $OH^-$ cannot be present in appreciable concentrations in a solution that is strongly acidic in nature.

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Water sits in an open beaker. assuming constant temperature and pressure, the vapor pressure of the water ______________ as the

user100 [1]

For a given system, vapor pressure can be defined as the pressure exerted by the vapor phase in equilibrium with the condensed phase.

In this case, as water sits in an open beaker it tends to evaporate thereby forming water vapor. This vapor phase then exerts a pressure over the liquid water, which is termed as the vapor pressure of water at that temperature and pressure. As water continues to evaporate, more and more molecules escape into the vapor phase which thereby increases the vapor pressure.

Ans: The vapor pressure of water increases as the water evaporates.

3 0

4 years ago

Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found

Igoryamba

Answer:

The concentrations are :

$[HAsc^-]=0.000702 M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

The pH of the solution is 3.15.

Explanation:

$H_2Asc\rightleftharpoons HAs^-+H^+$ $K_{a1}=1.0\times 10^{-5}$

Initial

c 0 0

Equilibrium

c-x x x

$K_{a1}=\frac{[HAs^-][H^+]}{[H_2Asc]}$

$1.0\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.0\times 10^{-5}=\frac{x^2}{(0.050-x)}$

Solving for x:

x = 0.000702 M

$[HAsc^-]=0.000702 M$

$HAsc^-\rightleftharpoons As^{2-}+H^+$ $K_{a2}=5\times 10^{-12}$

Initially

x 0 0

At equilibrium ;

(x - y) y y

$K_{a2}=\frac{[As^{2-}][H^+]}{[HAsc^-]}$

$5\times 10^{-12}=\frac{y\times y}{(x-y)}$

$5\times 10^{-12}=\frac{y^2}{(x-y)}$

Putting value of x = 0.000702 M

$5\times 10^{-12}=\frac{y^2}{(0.000702 -y)}$

$y=5.92\times 10^{-8} M$

$[Asc^{2-}]=5.92\times 10^{-8} M$

Total concentration of $[H^+]=x+y=0.000702 M+5.92\times 10^{-8} M=7.0206\times 10^{-4} M$

The pH of the solution :

$pH=-\log[H^+]$

$pH=-\log[7.0206\times 10^{-4} M}=3.15$

7 0

3 years ago

How much energy would be released as an electron moved from the n=4 to the n=3 energy level?

Thepotemich [5.8K]

The energy released when electron move from n=4 to n=3 is 0.66 eV

We know that in an atom energy of nth state is

$E_n = -13.6/n^{2}$ eV

where n is the energy level

Therefore,

$E_4 = -13.6/4^{2} \\E_3 = -13.6/3^{2}$

Thus, $E_4$ = -0.85eV

$E_3$ = -1.51eV

Therefore, total mount of energy released in moving electron from n=4 to n=3 is given by -

$E_4 -E_3$

= -0.85 - ( -1.51)

= 0.66eV

To know more about energy released in electron transition

brainly.com/question/8384785

#SPJ4

8 0

2 years ago

A catalyst will

scZoUnD [109]

Answer:

increase the chemical rate

7 0

3 years ago

The voltage generated by the zinc concentration cell described by the line notation Zn ( s ) ∣ ∣ Zn 2 + ( aq , 0.100 M ) ∥ ∥ Zn

Oxana [17]

Answer:

0.193 M

Explanation:

We need to calculate the Zn²⁺ concentration at the cathode where reduction occurs which is the right side in the expression:

Zn2+(aq,0.100 M) ‖ Zn2+(aq,? M) | Zn(s)

Zn²⁺ (aq,?) + 2 e⁻ ⇒ Zn (s)

and oxidation will occur in the anode

Zn (s) ⇒ Zn²⁺ (aq, 0.100 M ) + 2 e⁻

and the overall reaction is

Zn²⁺ (aq,?) ⇒ Zn²⁺ (0.100 M )

The driving force is the difference in concentration and E the electromotive force will be given by

E = Eº - 0.0592/2 log [Zn²⁺ (0.100 M) / Zn²⁺ (M) ]

Plugging the value for E and knowing Eº is cero because we have the same electrodes, we have

17.0 x 10⁻³ = 0 - 0.0592 log 0.100/ X =

- 17.0 x 10⁻³ / 0.0592 = log 0.100 / X

- 0.287 = log (0.100 / X)

Taking inverse log to both sides of the equation

0.516 = 0.100 / X ⇒ X = 0.100 / 0.516 = 0.193 M

4 0

3 years ago