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alekssr [168]
2 years ago
15

Approximately how many moles of Al3+ are reduced when 0.1 faraday of charge passes through a cell during the production of Al? (

Note: Assume there is excess Al3+ available and that Al3+ is reduced to Al metal only.)
A. 0.033 molB. 0.050 molC. 0.067 molD. 0.10 mol
Chemistry
1 answer:
Natali [406]2 years ago
6 0

Answer:

0.033 mol

Explanation:

1 faraday is require to  1 mol of electron

0.1 faraday is required for 0.1 mol of electron

3 mol of electron is required for 1 mole Al3+ to Al

0.1 mole of electron will therefore need ( 0.1 mol × 1 mol ) / 3 mol = 0.033 mol

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If charges flow very slowly through a metal, why does it not require several hours for a light to come on when you throw a switc
Harlamova29_29 [7]
T<span>he charges themselves may not move fast, but the force upon them does. The electric field set up by the battery or generator propagates through the wires at the speed of light.

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6 0
2 years ago
Which of the following compounds has polar covalent bonds: NaBr, Br2, HBr, and CBr4?
svetlana [45]

Answer: Option (e) is the correct answer.

Explanation:

A bond that is formed when an electron is transferred from one atom to another results in the formation of an ionic bond.

For example, NaBr will be an ionic compound as there is transfer of electron from Na to Br.

Whereas a bond that is formed by sharing of electrons is known as a covalent bond.

For example, CBr_{4} will be a covalent compound as there is sharing of electron between carbon and bromine atom.

Also, when electrons are shared between the combining atoms and there is large difference in electronegativity of these atoms then partial charges develop on these atoms. As a result, it forms a polar covalent bond.

For example, in a HBr compound there is sharing of electrons between H and Br. Also, due to difference in electronegativity there will be partial positive charge on H and partial negative charge on Br.  

Thus, we can conclude that out of the given options HBr is the only compound that has polar covalent bonds.

8 0
2 years ago
A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6
Alexxandr [17]
Specific heat capacity is the required amount of heat per unit of mass in order to raise teh temperature by one degree Celsius. It can be calculated from this equation: H = mCΔT where the H is heat required, m is mass of the substance, ΔT is the change in temperature, and C is the specific heat capacity.

H = m<span>CΔT
2501.0 = 0.158 (C) (61.0 - 32.0)

C = 545.8 J/kg</span>·°C
5 0
2 years ago
The standard curve was made by spectrophotographic analysis of equilibrated iron(III) thiocyanate solutions of known concentrati
posledela

Answer:

Molar concentration of the Fe³⁺ in the unknown solution is 8.01x10⁻⁵M.

Explanation:

When you make a calibration curve in a spectrophotographic analysis you are applying the Lambert-Beer law that states the concentration of a compound is directely proportional to its absorbance:

A = E*l*C

<em>Where A is absorbance, E is molar absorption coefficient, l is optical path length and C is molar concentration</em>

<em />

Using the equation of the line you obtain:

y = 4541.6X + 0.0461

<em>Where Y is absorbance and X is concentration -We will assume concentration is given in molarity-</em>

As absorbance of the unknown is 0.410:

0.410 = 4541.6X + 0.0461

X = 8.01x10⁻⁵M

<h3>Molar concentration of the Fe³⁺ in the unknown solution is 8.01x10⁻⁵M.</h3>

<em />

6 0
2 years ago
What is the value of ΔGo in kJ at 25 oC for the reaction between the pair: Mn(s) and Ag+(aq) to give Ag(s) and Mn2+(aq) Use the
Leni [432]

Answer:  -3.8\times 10^{5}J

Explanation:

Mn+2Ag^{+}\rightarrow Mn^{2+}+2Ag

Here Mn undergoes oxidation by loss of electrons, thus act as anode. silver undergoes reduction by gain of electrons and thus act as cathode.

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials.

E^0_{[Mn^{2+}/Mn]}= -1.18V

E^0_{[Ag^{2+}/Ag]}=+0.80V

E^0=E^0_{[Ag^{+}/Ag]}- E^0_{[Mn^{2+}/Mn]}

E^0=+0.80- (-1.18V)=1.98V

The standard emf of a cell is related to Gibbs free energy by following relation:

\Delta G^0=-nFE^0

\Delta G^0 = gibbs free energy

n= no of electrons gained or lost  = 2

F= faraday's constant

E^0 = standard emf  = 1.98V

\Delta G^0=-2\times 96500\times (1.98)=-3.8\times 10^{5}J

Thus the value of \Delta G^0 is -3.8\times 10^{5}J

8 0
2 years ago
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