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ss7ja [257]
3 years ago
12

The value of the equilibrium constant for the combination of nitrogen and oxygen to make NO is 2 x 10-9. What does this tell you

about the concentrations of the materials in the equilibrium mixture?
Group of answer choices

The concentration of reactants exceeds the concentration of products

The reactants are solids

The concentrations of the reactants and products are equal

The concentration of products exceeds the concentration of reactants
Chemistry
1 answer:
Rama09 [41]3 years ago
7 0

Answer:

A

Explanation:

A

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3) The total number of orbital’s in a shell with principle quantum no (n) is​
melamori03 [73]

Answer:

nine

There are nine orbitals in the n = 3 shell. There is one orbital in the 3s subshell and three orbitals in the 3p subshell. The n = 3 shell, however, also includes 3d orbitals. The five different orientations of orbitals in the 3d subshell are shown in the figure below.

Explanation:

8 0
3 years ago
Acid indigestion is sometimes neutralized with an antacid such as magnesium hydroxide (Mg(OH)2). What products will be released
NemiM [27]

Answer:

Magnesium chloride and water  

Explanation:

Mg(OH)₂ + 2HCl ⟶                MgCl₂                +   2H₂O

                                    magnesium chloride        water

7 0
3 years ago
Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --
frosja888 [35]

Answer :

(a) The anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

Thus, the anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}

The overall cell reaction will be,

2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}

E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V

E^o_{[Cu^{2+}/Cu]}=0.34V

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

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

E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}

E^o=1.54V-(0.34V)=1.20V

Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}

E_{cell}=1.022V

Therefore, the emf of cell potential is 1.022 V

5 0
3 years ago
Fill in the missing in information: (protons, electrons, atomic#, or
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Answer:

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3 0
2 years ago
What is the shape pf a molecule with a triple bond
makkiz [27]

Explanation:

A molecule with a triple bond will have a linear geometry as the angles between the bonds will be 180 degrees. Hence, it will be a linear geometry.

For example, a molecule of dinitrogen (N_{2}) has nitrogen atoms bonded through triple bond as follows.

                                      :N ≡ N:

Since valency of nitrogen is 5 therefore, each nitrogen will have one lone pair of electrons.

8 0
3 years ago
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