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Ray Of Light [21]
3 years ago
11

A voltaic cell has a zinc anode and a copper cathode. They are connected by a wire but no salt bridge. What can you predict will

happen? A. Without a salt bridge to slow it down the reaction will occur faster than normal. B. The electrons will flow to the zinc anode where a negative charge will build up and eventually halt the reaction. C. None of these D. The electrons will flow to the copper cathode where a negative charge will build up and eventually halt the reaction.
Chemistry
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:b

Explanation:

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The standard cell potential (Eºcell) for the reaction below is +0.63 V. The cell potential
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Valence electrons determine...(one word)
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Valence electrons determine its valency ? (The group, as in its oxidation state and stability)
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Consider the following reaction: 2H2S (g) + 3O2 (g)  2SO2 (g) + 2H2O (g) If O2 was the excess reagent, 8.3 mol of H2S were cons
Ganezh [65]

1. The balanced equation tells us that 2 moles of H2S produce 2 moles of H2O.  

8.3 moles H2S x (2 moles H2O / 2 moles H2S) = 8.3 moles H2O = theoretical amount produced  

8.3 moles H2O x (18.0 g H2O / 1 mole H2O) = 149 g H2O produced theoretically  

% yield = (actual amount produced / theoretical amount) x 100 = (137.1 g / 149 g) x 100 = 91.8 % yield  

2. Calculate moles of each reactant.  

150.0 g N2 x (1 mole N2 / 28.0 g N2) = 5.36 moles N2  

32.1 g H2 x (1 mole H2 / 2.02 g H2) = 15.9 moles H2  

The balanced equation tells us that we need 3 moles of H2 to react with every 1 mole of N2.  

So if we have 5.36 moles N2, we need 3x that = 16.1 moles H2. Do we have that much available? No, just under at 15.9 moles. So H2 is the limiting reactant. At the end of the reaction there will be a little N2 left over.

8 0
3 years ago
Read 2 more answers
In a laboratory experiment the reaction of 3.0 mol of H2 with 2.0 mol of I2 produced 1.0 mole of HI. Determine theoratical yield
Kobotan [32]

Answer:

Theoretical yield of HI is 512 g.

The percent yield for this reaction is 25%.

Explanation:

H_2+I_2\rightarrow 2HI

Moles of hydrogen gas = 3.0 moles

Moles of iodine gas  = 2.0 moles

According to reaction 1 mol of hydrogen gas reacts with 1 mol of iodine gas.

Then 3.0 moles of hydrogen gas reacts with 3.0 mol of iodine gas. But there are 2.0 moles of iodine gas. Hence,Iodine is a limiting reagent. The production of HI will depend upon iodine gas moles.

According to reaction , 1 mol of iodine gas gives 2 moles of HI.

Then 2 moles of iodine gas will give:

\frac{2}{1}\times 2 mol=2 mol of HI

Theoretically we will get 4 moles of HI.

Theoretical yield of HI =  4 mol × 128 g/mol= 512 g

Experimental yield of HI = 1.0 mol

= 1 mol × 128 g/mol= 128 g

\%yield=\frac{\text{Experimental yield}}{\text{Theoretical yield}} \times 100

\%yield=\frac{128 g}{512 g}\times 100=25\%

The percent yield for this reaction is 25%.

6 0
3 years ago
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