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jeyben [28]
3 years ago
7

An electrochemical cell contains a standard hydrogen electrode and a cathode consisting of a metallic chromium electrode, Cr(s),

in contact with a 1.00 M chromium solution, Cr3 (aq). The voltage produced by this cell was measured at 25oC. Which statements describe the results of this measurement, assuming the conditions are ideal? The cell voltage with the appropriate sign equals
I. the cell potential.
II. the electromotive force.
III. the standard cell potential.
IV. the standard reduction potential for Cr3+/Cr.

a. I only
b. l and Il
c. I, II and III
d. I, II, II, and IV
e. ll only
Chemistry
1 answer:
stira [4]3 years ago
5 0

Answer:

c. I, II and III

Explanation:

The cell is as follows

Cr / Cr⁺²(1M) // H⁺ ( 1 M ) / H₂

Standard reduction potential of hydrogen cell is zero . Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell

Cell potential = Ecathode - Eanode

= 0 - ( - E)

= E

E is cell potential and also standard cell potential or emf of the cell .

Standard reduction potential that is for Cr3+/Cr.  is  - E .

Hence statement I , II , III are right . IV th statement is wrong because of sign

Option c is correct.

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WHAT MASS OF WATER WILL BE PRODUCED FROM 2.70 MOLES OF CA(OH)2 REACTING WITH HCI
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<u>Explanation:</u>

We are given:

Moles of calcium hydroxide = 2.70 moles

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Ca(OH)_2+HCl\rightarrow CaCl_2+2H_2O

By Stoichiometry of the reaction:

1 mole of calcium hydroxide produces 2 moles of water

So, 2.70 moles of calcium hydroxide will produce = \frac{2}{1}\times 2.70=5.40mol of HCl

To calculate mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of water = 18 g/mol

Moles of water = 5.40 moles

Putting values in above equation, we get:

5.40mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(5.40mol\times 18g/mol)=97.2g

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3 0
4 years ago
How many miles can you drive, if your car gets 21.3 miles per gallon and you have 12.0 gallons of gas?
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255.6

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How many moles are in a solution with a concentration of 5 M and a volume of 0.25 L?
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Let, [Ag+] = [Cl-] = S

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Now, Molarity of solution = \frac{\text{weight of solute}}{\text{Molecular weight X volume of solution (l)}}
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