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LuckyWell [14K]
3 years ago
6

Enthalpy of reaction described for 3Mg+N2---->Mg3N2?

Chemistry
1 answer:
Finger [1]3 years ago
5 0

Oxidation-Reduction Reactions. The term oxidation was originally used to describe reactions in which an element combines with oxygen. Example: The reaction between magnesium metal and oxygen to form magnesium oxide involves the oxidation of magnesium.

Hope I helped :)

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How many electrons would be transferred in either a voltaic or electrolytic cell that uses the following half reactions
omeli [17]

Answer:

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

Explanation:

Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V

Mg^{2+}(aq) + 2 e^- \rightarrow Mg (s),E^o = -2.37 V

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

Reduction : cathode

Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V..[1]

Oxidation: anode

Mg(s)\rightarrow Mg^{2+}(aq) + 2 e^-,E^o = 2.37 V..[2]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

E^o_{cell}=-0.036V-(-2.37 V)=2.334 V

The overall reaction will be:

2 × [1] + 3 × [2] :

2Fe^{3+} (aq) + 3Mg(s)+6e^-\rightarrow 2Fe (s)+3Mg^{2+}(aq)+6e^-

Electrons on both sides will get cancelled :

2Fe^{3+} (aq) + 3Mg(s)\rightarrow 2Fe (s)+3Mg^{2+}(aq)

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

5 0
3 years ago
Identify the reactant and products in the following chemical reaction: CH4 + 2O2 ➡ CO2 + 2H2O
kap26 [50]

The reactants are methane and oxygen.

The products are carbon dioxide and water.

6 0
3 years ago
How many atoms of each element are in the chemical formula P2O5?
marshall27 [118]

Answer:A

Explanation:2 phosphorus and 2 oxygen

4 0
3 years ago
Help me with this question?!!​
olya-2409 [2.1K]

Answer:

500 mL

Explanation:

Step 1: Find conversions

1 mL = 0.0338 oz

Step 2: Use Dimensional Analysis

16.9 \hspace{2} oz(\frac{1 \hspace{2} mL}{0.0338 \hspace{2} oz} ) = 500 mL

3 0
3 years ago
Use the following balanced reaction to solve:
Naily [24]

Answer:  60.7 g of PH_3 will be formed.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given volume}}{\text{Molar volume}}    

\text{Moles of} H_2=\frac{60L}{22.4L}=2.68moles

The balanced chemical reaction is

P_4(s)+6H_2(g)\rightarrow 4PH_3(g)

H_2 is the limiting reagent as it limits the formation of product and P_4 is the excess reagent.

According to stoichiometry :

6 moles of H_2 produce = 4 moles of PH_3

Thus 2.68 moles of H_2 will produce=\frac{4}{6}\times 2.68=1.79moles  of PH_3

Mass of PH_3=moles\times {\text {Molar mass}}=1.79moles\times 33.9g/mol=60.7g

Thus 60.7 g of PH_3 will be formed by reactiong 60 L of hydrogen gas with an excess of P_4

3 0
3 years ago
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