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1 year ago

find the activity coefficient of h , in a solution containing 0.010 m hcl plus 0.040 kclo4. what is the ph of the solution?

Chemistry

1 answer:

Lisa [10]1 year ago

4 0

The activity coefficient of h is 0.05

The ph of the solution is 2.08

To find the activity coefficient, apply the concept of ionic equilibrium

h( activity coefficient)= 1/2CZ²

C is the concentration of species

Z is the charge on individual species

h= 1/2( 0.01 x1²+ 0.04x (-1)²+0.01 x1²+0.04x (-1)²)

h= 0.05

For the pH of the solution, apply the formula

pH= -log( H⁺x0.83)

pH = - log(0.01x0.83)

= 2.08

To learn more about the pH of the solution, visit brainly.com/question/26767292

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Answer:

lithium bromide... LiBr

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2 years ago

+ c) FeCl3 + NH4OH Fe(OH)3 NHACI

bixtya [17]

The question is incomplete, the complete question is:

Write the net ionic equation for the below chemical reaction:

(c): $FeCl_3+3NH_4OH\rightarrow Fe(OH)_3+3NH_4CI$

Answer: The net ionic equation is $Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)$

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

(c):

The balanced molecular equation is:

$FeCl_3(aq)+3NH_4OH(aq)\rightarrow Fe(OH)_3(s)+3NH_4Cl(aq)$

The complete ionic equation follows:

$Fe^{3+}(aq)+3Cl^-(aq)+3NH_4^+(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq)+3Cl^-(aq)$

As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)$

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3 years ago

H2SO4 +HI → __ H2S+12 +H2O balance the equation

Citrus2011 [14]

Answer:

H2SO4 + 8HI → H2S + 4I2 + 4H2O

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2 years ago

En un recipiente colocamos unos cubos de hielo luego tapamos y observamos que sucede despues de un tiempo

SSSSS [86.1K]

Three questions come along with the given statement. It is in Spanish language:

a) Por qué se humedeció la parte exterior del frasco?

b) Por qué el hielo disminuyó su volumen y ahora es agua?

c) Por qué puede haber agua en el exterior?

These are the three answers (in English).

First question:

a) Por qué se humedeció la parte exterior del frasco?

The question is Why did the outside of the bottle get wet?

Answer:

The outside of the bottle get wet because the ice cubes cooled the walls of the bottle, so the air surrounding the bottle also cooled.

The air contains humidity (water) in gas phase. The hotter the air the more the amount of humidity it can retain, the cooler the air the less the amount of humidity it can retain.

Then, when the air close to the walls of the bootle got cooler some of the water in the air became liquid and those are the drops of water that you see in the outside of the bottle.

Second question

b) Por qué el hielo disminuyó su volumen y ahora es agua?

The question is Why did the ice diminish its volume and now it is water?

Answer:

The ice diminished its volume and now it is water, becasue the ice, which is cooler than the surroundings, received heat energy (from the surroundings) and then its temperature increased. At some moment, this temperature reached the melting point of the ice (water) and it started to become liquid.

Third question

c) Por qué puede haber agua en el exterior?

The question is: Why can there be water outside?

Answer:

The water outside is outside since the beginning: it is in the air. You do not see it because it is gas state. When the air close to the walss of the bottle got cooler, part of the water in the air became liquid.

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3 years ago

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans

Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of $H_2SO_4$ (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

$H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of $H_2SO_4(aq)$ ? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of $H_2SO_4$ (aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_1,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL$

Putting values in above equation, we get:

$2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M$

0.044 M is the molarity of $H_2SO_4$ (aq).

4 0

3 years ago