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4 years ago

26 Which process occurs in an operating voltaic cell?(1) Electrical energy is converted to chemical energy.

Chemistry

2 answers:

tia_tia [17]4 years ago

8 0

Answer is: (2) Chemical energy is converted to electrical energy.

An electrochemical cell (voltaic or galvanic cell) is generating electrical energy from chemical reactions.

In galvanic cell, specie (for example zinc and zinc cations) from one half-cell, lose electrons (oxidation) and species from the other half-cell (for example copper and copper cations) gain electrons (reduction).

Oxidation on the zinc anode: Zn(s) → Zn²⁺(aq) + 2e⁻.

Reduction on the copper cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).

Alchen [17]4 years ago

7 0

The answer is (2) chemical energy is converted to electrical energy. The choice of (1) describes the electrolytic cell. The oxidation takes place at the anode and reduction takes place at the cathode.

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Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benz

stiv31 [10]

Answer: Freezing point of a solution will be $-1.16^0C$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(5.50-T_f)^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant = $5.12^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

$(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}$

$(5.50-T_f)^0C=6.68$

$T_f=-1.16^0C$

Thus the freezing point of a solution will be $-1.16^0C$

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3 years ago

If the number is not a proper coefficient, how would you make it one?

noname [10]

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2 years ago

Which of the following states that an orbital can contain two electrons only if all other orbitals at that sublevel contain at l

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How many grams of Br are in 335g of CaBr2

ASHA 777 [7]

The answer is 267.93 g

Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
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Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol

The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%

Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
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3 years ago

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bija089 [108]

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4 years ago