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grandymaker [24]
4 years ago
6

When silver nitrate is added to the Fe/SCN equilibrium, why is the colorless intense and a precipitate forms?

Chemistry
1 answer:
Reptile [31]4 years ago
3 0

Answer:

Here's what I get  

Explanation:

You have an equilibrium reaction between Fe³⁺/ SCN⁻ and FeSCN²⁺.

\underbrace{\hbox{Fe$^{3+}$}}_{\text{pale yellow-green}} +\underbrace{\hbox{SCN$^{-}$}}_{\text{colourless}} \, \rightleftharpoons \, \underbrace{\hbox{Fe(SCN)$^{2+}$}}_{\text{deep blood red}} \\

When you add AgNO₃, the Ag⁺ reacts with the SCN⁻. It forms a colourless precipitate of Ag(SCN).

Ag⁺(aq) + SCN⁻(aq) ⟶ AcSCN(s)

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

If you add Ag⁺ to the equilibrium solution, it removes the SCN⁻ [as an Ag(SCN) precipitate].

The system responds by trying to replace the missing SCN⁻:

The Fe(SCN)²⁺ dissociates to form SCN⁻, so the position of equilibrium shifts to the left,

You now have more Fe³⁺ and SCN⁻ and less of the highly coloured Fe(SCN)²⁺ at the new equilibrium.

The deep red colour becomes less intense.

 

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An electric range burner weighing 699.0 grams is turned off after reaching a temperature of 482.0°C, and is allowed to cool down
jasenka [17]

Answer:

0.42 J/gºC

Explanation:

We'll begin by calculating the heat energy used to heat up the water. This can be obtained as follow:

Mass (M) of water = 560 g

Initial temperature (T₁) = 22.7 °C

Final temperature (T₂) = 80.3 °C.

Specific heat capacity (C) of water = 4.18 J/gºC

Heat (Q) absorbed =?

Q = MC(T₂ – T₁)

Q = 560 × 4.18 (80.3 – 22.7)

Q = 2340.8 × 57.6

Q = 134830.08 J

Finally, we shall determine the specific heat capacity of the burner. This can be obtained as follow:

Mass (M) of burner = 699 g

Initial temperature (T₁) = 482.0°C

Final temperature (T₂) = 22.7 °C

Heat (Q) evolved = – 134830.08 J

Specific heat capacity (C) of the burner =?

Q = MC(T₂ – T₁)

–134830.08 = 699 × C (22.7 – 482.0)

–134830.08 = 699 × C × –459.3

–134830.08 = –321050.7 × C

Divide both side by –321050.7

C = –134830.08 / –321050.7

C = 0.42 J/gºC

Therefore, the specific heat capacity of the burner is 0.42 J/gºC

8 0
3 years ago
In an attempt to conserve water and to be awarded LEED (Leadership in Energy and Environmental Design) certification, a 20,000-l
kondaur [170]

Answer:

rate of water condensation in cistern = 2,604.628L/ min

the hours of operation required to fill the cistern 0.128hr

Explanation:

Given,

At 22°C, the properties of conditioned air are-

Flowrate = 2830m^3/min ; [1 m^3= 1000 L]

= 2830 x (1000L) / min

= 2.830 x 10^6 L

Consider:

intake at 31°C = X liters/ min.

Therefore

X liters = volume of air flowing per minutes

Moisture content (relative humidity)

= 70.0 % of X L = 0.70X L

Dry (some moisture removed) air content

= X L - 0.70X L = 0.30 L

Used charles' law to determine the Volume of released air at 31°C

(V1/ T1) = (V2/ T2) - equation 1

Where,

V1 = 2830m^3

T1 = 22°c = 295k

V2 = ?

T2 = 31°c = 304k

2830m^3/ 295k = V2/304k

V2 = 2830m^3 × 304k

--------------------------

295k

= 2916.339m^3

Therefore,

The volume of same exaled air (2830m^3/min at 22°c) is equal to 2916.339m^3 at 31°c

During condensation, only water is removed

Therefore

The volume of dry gas and 50% relative humidity is equal to 2916.339m^3 at 22°c

Now, only water is removed during condensation. After removing water, the volume of dry gas and 40% RH is equal to 2916.339m^3

So,

Dry air content + 50% of dry air content

=0.30xm^3 + 50% of 0.30xm^3

=0.30xm^3 + 0.15xm^3

= 0.45xm^3

Intake per minute = x = ?

Let

0.45xm^3 = 2916.339m^3

X = 2916.339m^3

-----------------------

0.45

X = 6480.754m^3

Therefore, intake per minute at 31°C = 6480.754m^3

Volume of moisture removed = volume of intake air at 31°C - Volume of exhausted air at 31°C

= 6480.754m^3 - 2916.339m^3

= 3,564.415m^3

Convert m^3 to L (1m^3 =1000L)

= 3,564.415 × 10^3L

= 3.564 × 10^6L

Assuming moisture behave ideally at 27°C, calculate the moles of water vapor using ideal gas equation-

PV = nRT ....equation 2

Where,

P = pressure in atm = 1.00atm

V = volume in L = 3.564 × 10^6L

n = number of moles = ?

R = universal gas constant= 0.0821 atm L mol-1K-1

T = absolute temperature (in K) = 300k

Putting the values for amount of moisture removed at 22°C-

1.00 atm x 3.564 × 10^6L = n x (0.0821 atm L mol-1K-1) x 300 K

n = 3.564 × 10^6atmL/ 24.63 atm L mol-1

n = 144,701.583 mol

Thus,

during conditioning, 144,701.583 mol of water was removed.

Mass of water removed = moles x molar mass

=144,701.583 mol x (18.0 g/ mol)

= 2,604,628.44g

= 2,604.62844 kg

= 2,604.628kg

Let density of water be 1.00 kg/ L at through the temperatures (31°C to 22°C), the volume of liquid water condensed in the cistern is given by-

Volume of water condensed = Mass of moisture removed x density of water

= 2,604.628kgx (1.00 kg/ L)

= 2,604.628L

A. Therefore, rate of water condensation in cistern = 2,604.628L/ min

B. Time required to fill the cistern = Capacity of cistern/ rate of water condensation

Given

Capacity of cistern = 20000L

= 20000 L/ (2,604.628L/ min)

= 7.679min

Equivalent to 0.128hr

5 0
3 years ago
Which best describes a mixture?
mixer [17]
A mixture means two or more substances forming into one new substance, look at your answers that you have. Does it show two or more different substances that chemically change into one new substance?
8 0
3 years ago
Read 2 more answers
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Virty [35]
Spinning top, moving car, and rolling ball have kinetic energy I believe
5 0
4 years ago
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Feliz [49]

Answer:

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Explanation:

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6 0
3 years ago
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