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OverLord2011 [107]
2 years ago
6

Calculate the amount of gold deposited when a current of 5A is passed through a solution of gold salt for 2hrs 15mins. If the sa

me current is used, find the time taken for 6g of gold to be deposited. (Au=197 1faraday=96500c)
Chemistry
1 answer:
Nataliya [291]2 years ago
8 0

Answer:

a. 82.68 g b. 9.8 min

Explanation:

a. The amount of gold deposited by 5 A current in 2 hrs 15 mins

Since charge Q = It where I = current = 5 A and time = 2 hrs 15 mins = 2 × 60 min + 15 min = 120 + 15 min = 135 min = 135 × 60 s = 8100 s

Q = 5 A × 8100 s

= 40500 C

Also, Q = nF where n = number of moles of gold deposited and F = Faraday's constant = 96500 C

n = Q/F = 40500 C/96500 C = 0.4195 moles ≅ 0.42 mole

Now n = m/M where m = mass of gold and M = molar mass of gold = 197

m = nM

= 0.42 × 197 g

= 82.68 g

b. The time taken for 6g of gold to be deposited.

We first find the number of moles of gold in 6g of gold

Since n = m/M and m = 6 g

n = 6/197 = 0.0305 mole

Q = It = nF

t = nF/I

= 0.0305 mol × 96500 C/5 A

= 2939.09 mol C

= 587.82 s

Changing t to minutes

587.82/60 s = 9.8 min

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570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students
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2 years ago
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A raindrop has a mass of 50. mg and the Pacific Ocean has a mass of 7.08 x 10^20 kg. what is the mass of 1 mole of raindrops? Ho
Anna007 [38]

Solution:

According to the Avogadro's number:  

6.022 *10^23 drops per 0.050 g/ drop = 3.011 *10^22 grams per mole of drops  

3.011 *10^22 grams per 1 kg / 1000 grams = 3.011 *10^19 kilograms / mole of drops  

thus the answers are:  

3.0 *10^22 grams per mole of drops  

3.0 *10^19 kilograms per mole of drops  

And,

In the calculation of how many moles of raindrops in the Pacific Ocean is:  

7.08X10^20kg per 3.0 *10^19 kilograms per mole of drops = 23.5 moles of drops

This is the required solution.  


7 0
2 years ago
Read 2 more answers
Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°
Pavel [41]

Answer:

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that  involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

PV = \frac{m}{M}RT

PxM = \frac{m}{V}RT

PxM = ρRT

ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

2.7 = \frac{0.94xMavg}{0.082x289}

0.94Mavg = 63.9846

Mavg = 68.0687 g/mol

The molar mass of N is 14 g/mol and of O is 16 g/mol, than M_{NO2} = 46 g/mol and M_{N2O4} = 96 g/mol. Calling y the molar fraction:

Mavg = M_{NO2}y_{NO2} + M_{N2O4}y_{N2O4}

And,

y_{NO2} + y_{N2O4} = 1

y_{N2O4} = 1 - y_{NO2}

So,

68.0687 = 46y_{NO2} + 92x(1 - y_{NO2})

68.0687 - 92 = 46y_{NO2} - 92y_{NO2}

46y_{NO2} = 23.9313

y_{NO2} = 0.52

y_{N2O4} = 0.48

The partial pressure is the molar fraction multiplied by the total pressure so:

PNO₂ = 0.52x0.94 = 0.49 atm

PN₂O₄ = 0.48x0.94 = 0.45 atm

8 0
2 years ago
A chamber with a fixed volume is shown above. The temperature of the gas inside the chamber before heating is 25.2 C and it’s pr
rusak2 [61]

Answer:

Explanation:

Given parameters:

Initial temperature T₁  = 25.2°C  = 25.2 + 273  = 298.2K

Initial pressure  = P₁  = 0.6atm

Final temperature = 72.4°C   = 72.4 + 273  = 345.4K

Unknown:

Final pressure = ?

Solution:

To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

                  \frac{P_{1} }{T_{1} }   = \frac{P_{2} }{T_{2} }

where P and T are temperatures, 1 and 2 are initial and final temperatures.

 Input the parameters and solve;

          \frac{0.6}{298.2}   = \frac{P_{2} }{345.4}  

          P₂   = 0.7atm

         

3 0
2 years ago
Hafnium has six naturally occurring isotopes: 0.16% of 174Hf, with an atomic weight of 173.940 amu; 5.26% of 176Hf, with an atom
ser-zykov [4K]

Answer:

177.277amu

Explanation:

the total occuring isotopes for Hafnium is =6.

First isotope had an atomic weight of 173.940amu

Second isotope =175.941amu

Third isotope =176.943amu

Fourth isotope=177.944amu

Fifth isotope. =178.946amu

sixth isotope .179.947amu

<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>

Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu

Average atomic weight= 1063.661amu /6 = 177.2768333amu

= 177.277amu to 3 decimal places.

6 0
2 years ago
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