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Snezhnost [94]
3 years ago
14

Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction. Ag+(aq)+e−→Ag(s) Part A Silver can be e

lectroplated at the cathode of an electrolysis cell according to the following half-reaction: Ag+(aq)+e−→Ag(s) What mass of silver plates onto the cathode when a current of 8.5 A flows through the cell for 68 min ?
Chemistry
1 answer:
dmitriy555 [2]3 years ago
3 0

Answer: Mass of silver deposited at the cathode is 37.1g

Explanation: According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode (cathode or anode) is directly proportional to quantity of electricity passed through the electrolyte

Faraday has found that to liberate one gm eq. of substance from an electrolyte, 96500C of electricity is required.

Ag^+ +e− ==> Ag(s)

Given that

Current (I) = 8.5A

Time (t) = 65 *60 = 3900s

Quantity of electricity passed = 8.5*3900 =33150C

Molar mass of Ag= 108g

96500C will liberate 108g

33150C will liberate Xg

Xg= (108*33150)/96500

=37.1g

Therefore the mass of Ag deposited at the cathode is 37.1g.

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An ideal gas in a closed container initially has a volume V and Temperature T the final tempera is 5/4T and the final pressure i
Anettt [7]

Answer:

V_2 = \frac{5V}{8}

Explanation:

I am assuming you are saying what is the final volume of the gas

Known :

Initial volume (V1) = V

Initial temperature (T1) = T

Final temperature (T2) = 5/4 T

Initial pressure (P1) = P

Final pressure (P2) = 2P

<u>Wanted: Final volume (V2)</u>

<u />

<u />\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\frac{PV}{T} = \frac{(2P)V_2}{(5/4)T}\\\frac{V}{1} = \frac{(2)V_2}{5/4}\\5/4V = 2V_2\\\\V_2 = \frac{5V}{8}

5 0
2 years ago
The common titanium alloy known as T-64 has a composition of 90 weight% titanium 6 wt% aluminum and 4 wt% vanadium. Calculate th
Anna007 [38]

Explanation:

Suppose in 100 g of alloy contains 90% titanium 6% aluminum and 4% vanadium.

Mass of titanium = 90 g

Moles of titanium = \frac{90 g}{47.87 g/mol}=1.8800 mol

Total number of atoms of titanium ,a_t=1.8800 mol\times N_A

Mass of aluminum = 6 g

Moles of aluminium = \frac{6 g}{26.98 g/mol}=0.2223 mol

Total number of atoms of aluminium,a_a=0.2223 mol\times N_A

Mass of vanadium  = 4 g

Moles of vanadium= \frac{4 g}{50.94 g/mol}=0.0785 mol

Total number of atoms of vanadiuma_v=0.0785 mol\times N_A

Total number of atoms in an alloy = a_t+a_a+a_v

Atomic percentage:

Atomic\%=\frac{\text{total atoms of element}}{\text{Total atoms in alloy}}\times 100

Atomic percentage of titanium:

:\frac{a_t}{a_t+a_a+a_v}\times 100=\frac{1.8800 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=86.20\%

Atomic percentage of Aluminium:

:\frac{a_a}{a_t+a_a+a_v}\times 100=\frac{0.2223 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=10.19\%

Atomic percentage of vanadium

:\frac{a_v}{a_t+a_a+a_v}\times 100=\frac{0.0785 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=3.59\%

6 0
4 years ago
A sodium ion has a radius of 1.16 x 10^-10 m and a nearby fluoride ion has a radius of 1.9 x 10^-10 m. Determine the distance be
Ann [662]

The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.

r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.

r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.

d(NaF) = r(Na⁺) + r(F⁻).

d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.

d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.

The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.

4 0
4 years ago
Suppose you are a chemical engineer working with a client who produces toothpaste.The client wants to add a blue stripe with min
Nikitich [7]

Answer:

Explanation:

I would ask him why would he want to add the blue stripe?

5 0
3 years ago
Please Answer Correctly
mylen [45]

Explanation: The radius of an object is found from the center of the object to the perimeter. Radius can be any number, but it is the measurement from the center to the perimeter.

4 0
3 years ago
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