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goldfiish [28.3K]
3 years ago
8

Using the two cell reduction potentials shown for their corresponding reaction, calculate the cell potential for a voltaic cell

made from these two systems. Question 16 options: A) 1.68 V B) –1.68 V C) –0.78 V D) 0.78 V
Chemistry
1 answer:
guajiro [1.7K]3 years ago
6 0

Answer:

The right alternative is Option D (0.78 V).

Explanation:

According to the question,

The cell potential will be:

= E^0_{Cr_2 O_7/er^{3+}}- E^0_{Fe^{2+}/Fe}

By putting the values, we get

= 1.23-0.45

= 0.78 \ V

Thus the above is the correct option.

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How many moles of phosphoric acid would be needed to produce 15 grams of water?
bekas [8.4K]

Moles of phosphoric acid would be needed : 0.833

<h3>Further explanation</h3>

Given

15 grams of water

Required

moles of phosphoric acid

Solution

Reaction(decomposition) :

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mol water (H2O :

= mass : MW

= 15 g : 18 g/mol

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7 0
3 years ago
Determine moles of 1.5g of sodium carbonate.
KIM [24]

Answer:

0.014mol

Explanation:

Given parameters:

Mass of Na₂CO₃  = 1.5g

Unknown:

Number of moles  = ?

Solution:

Number of moles of a compound is mathematically expressed as;

  Number of moles  = \frac{mass}{molar mass}  

 Molar mass of  Na₂CO₃  = 2(23) + 12 + 3(16) = 106g/mol

 Number of moles = \frac{1.5}{106}   = 0.014mol

8 0
3 years ago
What product of an acid base reaction is an ionic compound apex?
charle [14.2K]
When acids react with bases they produce salt and water such as:
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According to strength of acid and base, we have 4 types of salts:
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salt of weak acid and weak base like: CH₃COONH₄
8 0
3 years ago
Read 2 more answers
In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit
natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.  

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,  

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.  

7 0
3 years ago
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