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timurjin [86]
3 years ago
10

When an aqueous solution of AgNO3 is electrolyzed, a gas is observed to form at the anode. The gas is

Chemistry
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

Oxygen

Explanation:

First, list out all the ions in the aqueous solution:

Ag+, NO3-

H+, OH-

In the anode, the substance lose electrons to undergo oxidation.

From the 4 ions, only OH- can lose electrons to form water and oxygen,

4OH- --->  O2 + 2H2O + 4e-

While others tend to gain electrons to form new substances instead (they undergo reduction).

Oxygen is the gas produced.

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A sample of 87.6 g of carbon is reacted with 136 g of
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Answer:

A. fluorine, 1.79 moles

Explanation:

Given parameters:

Mass of carbon  = 87.7g

Mass of fluorine gas  = 136g

Unknown:

The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced  = ?

Solution:

   Equation of the reaction:

             C    +   2F₂ →   CF₄  

let us find the number of the moles the given species;

  Number of moles = \frac{mass}{molar mass}  

  C;   molar mass = 12;

            Number of moles  = \frac{87.7}{12}   = 7.31moles

 F;  molar mass  = 2(19)  = 38g/mol

             Number of moles  = \frac{136}{38}   = 3.58moles

 So;

   From the give reaction:

          1 mole of C requires 2 moles of F₂

         7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles

But we have 3.58 moles of the F₂;

  Therefore, the reactant in short supply is F₂ and it is the limiting reactant;

 So;

       2 moles of F₂ will produce  mole of CF₄  

       3.58 moles of F₂ will then produce \frac{3.58}{2}  = 1.79moles of CF₄

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Luba_88 [7]

Answer:

Explanation:

1)

Given data:

Mass of lead = 25 g

Initial temperature = 40°C

Final temperature = 95°C

Cp = 0.0308 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 95°C -  40°C

ΔT = 55°C

Q = 25 g × 0.0308 j/g.°C  × 55°C

Q = 42.35 j

2)

Given data:

Mass  = 3.1 g

Initial temperature = 20°C

Final temperature = 100°C

Cp = 0.385 j/g.°C

Heat required = ?

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 100°C -  20°C

ΔT = 80°C

Q = 3.1 g × 0.385 j/g.°C  × 80°C

Q = 95.48 j

3)

Given data:

Mass of Al = ?

Initial temperature = 60°C

Final temperature = 30°C

Cp = 0.897 j/g.°C

Heat released = 120 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

ΔT = 30°C -  60°C

ΔT = -30°C

120 j = m × 0.897 j/g.°C  × -30°C

120 j = m × -26.91  j/g

m = 120 j / -26.91  j/g

m =  4.46 g

negative sign show heat is released.

4)

Given data:

Mass of ice = 1.5 g

Change in temperature  = ?

Cp = 0.502 j/g.°C

Heat added= 30.0 j

Solution:

Specific heat capacity: Cp

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature = initial temperature

30.0 j = 1.5 g × 0.502 j/g.°C  × ΔT

30.0 j = 0.753 j/°C  × ΔT

30.0 j /0.753 j/°C  = ΔT

39.84 °C  =  ΔT

3 0
3 years ago
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