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valentinak56 [21]

3 years ago

9

When aqueous solutions of NH4OH(aq) and CuCl2(aq) are mixed, the products are NH4Cl(aq) and Cu(OH)2(s). What is the net ionic eq

uation for this reaction

1 answer:

Ilia_Sergeevich [38]3 years ago

5 0

Answer:

2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)

Explanation:

The net ionic equation usually shows the main ionic reaction that goes in the system. The other ions that do not participate in this net ionic equation are called spectator ions. Spectator ions do not participate in the main reaction occurring in the system.

The net ionic equation quite often result in the formation of a solid precipitate in the system such as Cu(OH)2.

The net ionic equation for this reaction is;

2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

djverab [1.8K]

Answer:

The new partial pressures after equilibrium is reestablished for $PCl_3$ :

$P_1'=6.798 Torr$

The new partial pressures after equilibrium is reestablished $Cl_2$ :

$P_2'=26.398 Torr$

The new partial pressures after equilibrium is reestablished for $PCl_5$ :

$P_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=P_1=13.2 Torr$

Partial pressure of the $Cl_2=P_2=13.2 Torr$

Partial pressure of the $PCl_5=P_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{P_1}{P_1\times P_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=P_1'=13.2 Torr$

Partial pressure of the $Cl_2=P_2'=?$

Partial pressure of the $PCl_5=P_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=P_1'+P_2'+P_3'$

$263.0Torr=13.2 Torr+P_2'+217.0 Torr$

$P_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{P_3'}{P_1'\times P_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished for $PCl_3$ :

$P_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

The new partial pressures after equilibrium is reestablished $Cl_2$ :

$P_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

The new partial pressures after equilibrium is reestablished for $PCl_5$ :

$P_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

5 0

2 years ago

Why is it appropriate to say the constitution is designed to "bend like a willow?

olga nikolaevna [1]

The Constitutions was designed to be flexible so new laws can be added as needed and old laws can be reinterpreted

3 0

2 years ago

A compound that conducts an electric current in aqueous solution or in the molten state.

makvit [3.9K]

Answer:

<em>A compound that conducts an electric current in aqueous solution or in the molten state is</em> a <u>ionic compound</u>.

E.g. NaCl (table salt)

Explanation:

Some examples of ionic compounds are NaCl, KF, KI, MgSO₄.

Ionic compounds are formed by positivie ions (named cations) and negative ions (named anions). The strong electrostatic atraction between cations and anions permits the formation of crystals, which are stuctures characterized by a regular pattern. In solid state the ions are in fixed positions.

In order to conduct electricity, the charged particles (cations and anions in the case of the ionic compounds) need to move freely.

Hence, since in solid state, ionic compounds do not conduct electric current. But, <em>in molten (liquid) state or in aqueous solution, ionic compounds conduct electric current because, then, the ions move freely.</em>

7 0

2 years ago

Calculate the number of grams nitric acid (HNO3) in 375 L of a 1.00 M solution.

Mrac [35]

You take the 1.00 M and multiply it by 375L witch gives you 375mol and then you take the mol and multiply it by the molar mass giving you 23630.505g nitric acid or HNO3

6 0

2 years ago

How many molecules are there in 237 grams of CCl4

Otrada [13]

Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We calculate as follows:

237g CCl4 (1 mol / 153.81 g ) (6.022 x 10^23 molecules / 1 mol ) = 9.28x10^23 molecules

3 0

3 years ago