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bekas [8.4K]
4 years ago
5

Which equations represent precipitation reactions?

Chemistry
2 answers:
Amanda [17]4 years ago
7 0

Answer : The correct option is, AgNO_3+NaCl\rightarrow AgCl+NaNO_3

Explanation :

Precipitation reaction : It is defied as the chemical reaction in which the two soluble salts solutions combined to form an insoluble salts.

(1) Na_2S+FeBr_2\rightarrow 2NaBr+FeS

(2) MgSO_4+CaCl_2\rightarrow MgCl_2+CaSO_4

(3) LiOH+NH_4I\rightarrow LiI+NH_4OH

(4) 2NaCl+K_2S\rightarrow Na_2S+2KCl

(5) AgNO_3+NaCl\rightarrow AgCl+NaNO_3

In the given reactions 1, 2, 3 and 4, the salt forms are completely soluble in water.

As we know that all the sodium, potassium and ammonium salts are soluble in water.

While in the reaction 5, the salt form that is silver chloride (AgCl) are insoluble in water. So, this reaction is a precipitation reaction.

Hence, the correct option is, AgNO_3+NaCl\rightarrow AgCl+NaNO_3

pychu [463]4 years ago
6 0

A. : In this reaction one of the product, FeS is insoluble. Therefore, this is a precipitation reaction.


B. : In this reaction, the product is a solid(insoluble). So, this is a precipitation reaction too.


C.: In this reaction, both the products are soluble. So this is not a precipitation reaction.


D.: In this reaction, both the products are soluble. So this is not a precipitation reaction.


E. : In this reaction, the product AgCl is a precipitate. So, it is a precipitation reaction.

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If 0.500 mol neon at 1.00 atm and 273 K expands against a constant external pressure of 0.100 atm until the gas pressure reaches
trapecia [35]

Answer:

The work done on neon = -323 J

The internal energy change= -392.84 J

The heat absorbed by neon = -69.84 J

Explanation:

Step 1: Data given

Number of moles  = 0.500 moles

Pressure  = 1 atm

Temperature  = 273 Kelvin

The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.

a) calculate the work done on neon

W = -P(V2-V1)    

⇒ with P = the pressure = 0.1 atm

⇒ with V1 = the initial volume = nRTi /Pi

⇒ with V2 = the final volume = nRTf /Pf

W = -PnR((T2/P2) -(T1/P1))

⇒ with T2 = the final temperature = 210 K

 ⇒ with T1 = the initial temperature = 273 K

 ⇒ with P2 = the final pressure = 0.200 atm

 ⇒ with P1 = the initial pressure = 1.00 atm

W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))

W = -nR*(105 - 27.3)

W= -(0.500)*(8.314)*(77.7)

W = -323 J

b) calculate the internal energy change

E = (3/2)*nRT

ΔE = Ef - Ei

ΔE =(3/2)*nR(T2-T1)

⇒ with n= number of moles = 0.500 moles

⇒ with T2 =the final temperature = 210 K

⇒ with T1 = the initial temperature = 273 K

ΔE = (3/2)*(0.5)*(8.314)(210-273)

ΔE = -392.84 J

c) Calculate the heat absorbed by neon

ΔE = q + W

q = ΔE -W

⇒ with ΔE = -392.84 J

⇒ with W = -323 J

q = -392.84 J -( -323 J)

q =-392.84 J + 323 J

q = -69.84 J

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9.00g/1hr * 1kg/100g * 1hr/60min = 0.00015kg/min or 1.5 * 10^-4kg/min.
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