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bekas [8.4K]
3 years ago
11

The reason H2S exists as a gas while H2O exists as a liquid at room temperature is because:

Chemistry
1 answer:
Masja [62]3 years ago
7 0
<span>The reason H2S exists as a gas while H2O exists as a liquid at room temperature is because H2O possesses hydrogen bonding, while H2S does not.
I'd say that the last option is correct.
</span>
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Which of the following examples describes an increase in potential energy? Question 1 options: a basketball falling through a ne
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Paint being carried up a ladder
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An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t
erik [133]

Answer:

\% Fe^{+2}=70%

Explanation:

Hello,

In this case, we could considering this as a redox titration:

Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}

Thus, the balance turns out (by adding both hydrogen ions and water):

Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-

Thus, by stoichiometry, the grams of Fe+2 ions result:

m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}

Finally, the mass percent is:

\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\  \% Fe^{+2}=70%

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8 0
2 years ago
What is oxidized in a galvanic cell with aluminum and gold electrodes ?
k0ka [10]

Answer:

Aluminum metal

Explanation:

In order to properly answer this or a similar question, we need to know some basic rules about galvanic cells and standard reduction potentials.

First of all, your strategy would be to find a trusted source or the table of standard reduction potentials. You would then need to find the half-equations for aluminum and gold reduction:

Al^{3+}+3e^-\rightarrow Al; E^o=-1.66 V

Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Since we have a galvanic cell, the overall reaction is spontaneous. A spontaneous reaction indicates that the overall cell potential should be positive.

Since one half-equation should be an oxidation reaction (oxidation is loss of electrons) and one should be a reduction reaction (reduction is gain of electrons), one of these should be reversed.

Thinking simply, if the overall cell potential would be obtained by adding the two potentials, in order to acquite a positive number in the sum of potentials, we may only reverse the half-equation of aluminum (this would change the sign of E to positive):Al\rightarrow Al^{3+}+3e^-; E^o=1.66 V\\Au^{3+}+3e^-\rightarrow Au; E^o=1.50 V

Notice that the overall cell potential upon summing is:

E_{cell}=1.66 V + 1.50 V=3.16 V

Meaning we obey the law of galvanic cells.

Since oxidation is loss of electrons, notice that the loss of electrons takes place in the half-equation of aluminum: solid aluminum electrode loses 3 electrons to become aluminum cation.

6 0
2 years ago
Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).
wolverine [178]

Answer: Li is the reducing agentg and O is the oxidizing agent.

Explanation:

1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.

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Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.

O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.

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Substituting the values:

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