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3 years ago

How many grams of water will form if 10.54 g H2 react with 95.10 g O2?

Chemistry

2 answers:

sergiy2304 [10]3 years ago

5 0

Answer:

94.0 g H2O

Explanation:

uranmaximum [27]3 years ago

5 0

Answer:94.0 g H2O

Explanation:

2H2 + O2 --> 2H2O

The

question asked me how many grams of water will form if 10.54 g H2 react with 95.10 g O2.

The correct answer was 94.0g of H2O I got from

10.54g H2 * 1 mol H2/2.02 g/mol H2 * 2 mol H2O/2 mol H2 = 5.22 mol H2O

95.10g O2* 1 mol O2/32 g/mol O2 * 2 mol H2O/1 mol O2 = 5.94 mol H2O

The limiting reactant is 5.22 mol, which 5.22 mol H2O * 18.01 g/mol H2O = 94.0g H2O.

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At high pressures how does the volume of a real gas compare with the volume of an ideal gas under the same conditions and why

malfutka [58]

Answer: No, at high pressures, volume of a real gas does not compare with the volume of an ideal gas under the same conditions.

Reason:
For an ideal gas, there should not be any intermolecular forces of interaction. However, for real gases there are intermolecular forces of interaction like dipole-dipole and dipole-induced dipole. Further, at high pressures, molecules are close by. Hence, extend of these intermolecular forces is expected to be high. This results in decreases in volume of real gas. Thus, volume of a real gas does not compare with the volume of an ideal gas under the same conditions.

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3 years ago

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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

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3 years ago

Question 8 Review Which metal jis most easily oxidized?

tester [92]

Answer:

The order of some common metals in the electromotive series, starting with the most easily oxidized, is: lithium, potassium, calcium, sodium, magnesium, aluminum, zinc, chromium, iron, cobalt, nickel, lead, hydrogen, copper, mercury, silver, platinum, and gold.

Explanation:

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mole ratios of hydrazine should be 1:2. I could be wrong. Are there any options to choose from?

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3 years ago

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Answer:

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3 years ago