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ikadub [295]
4 years ago
8

An electrochemical cell is powered by the half reactions shown below.

Chemistry
1 answer:
andrezito [222]4 years ago
7 0
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation. 

To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign. 

Pb(s) --> Pb2+ +2e-      E0 = +0.13 V
Ag+ + e-  ---> Ag           E0 = +0.80 V

Adding up the E0's would yield an overall electric cell potential of +0.93 V.
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How much heat is absorbed/released when 20.00 g of NH3(g) (17.0307g/mol) react in the presence of excess O2(g) to produce NO(g)
irina1246 [14]

Answer:

342.8 kJ are absorbed

Explanation:

In the reaction:

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) ΔH° = 1168 kJ

<em>As ΔH > 0, the heat is absorbed. Also, when 4 moles of NH3 are involved in the reaction, there are absorbed 1168 kJ</em>.

Having this in mind, moles of NH3 in 20.00g are:

20.00g × (1mol / 17.0307g) = <em>1.174 moles</em>

<em></em>

Thus, 1.174 moles of NH3 absorbed:

1.174 moles × (1168 kJ / 4 moles) = <em>342.8 kJ are absorbed</em>.

8 0
4 years ago
If an ice cube has a mass of 25 grams and it melts, what would the mass of the liquid water be?
sesenic [268]

Answer:25grams

Explanation: the mass stays the same even when it changes form

6 0
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What is alcohol made from?
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5 0
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The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How
viktelen [127]
The exact molecular mass for butane (C4H10) is 
12.0096*4+1.0079*10=58.1174  which is 58.1 to 3 significant figures.

Proportion of carbon in the compound
12.0096*4: 58.1174
=>
48.0384 : 58.1174

The mass of carbon in 2.50 grams of butane can be obtained by proportion, namely
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= 2.50 * (48.0384/58.1174)
= 2.0664
= 2.07 g (approximated to 3 significant figures)

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3 years ago
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