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3 years ago

What happens if more product is added to a system at equilibrium apex?

1 answer:

6 0

According to Le Chatelier's Principle," If a system (reaction) at equilibrium is subjected to any external stress, either by changing pressure, temperature or concentration of reactants or products, the system will adjust itself in such a way that the effect is minimized".

So, when more product is added to reaction at equilibrium, the equilibrium will shift in backward direction in order to minimize the effect of concentration of product.

Result:
With the addition of product, the rate of backward reaction will increase resulting in increasing the concentration of reactants.

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3 years ago

A microwave operates at 7.42 mm wavelength. How much energy is produced in Joules from 359 photons?

alisha [4.7K]

Answer:

$E=9.62\times 10^{-21}\ J$

Explanation:

Given that,

The wavelength of a microwave is 7.42 mm or 0.00742 m

No. of photons, n = 359

We need to find the energy produced by this no of photons. It can be given by the formula as follows :

$E=\dfrac{nhc}{\lambda}\\\\E=\dfrac{359\times 6.63\times 10^{-34}\times 3\times 10^8}{0.00742}\\\\=9.62\times 10^{-21}\ J$

$E=9.62e^{-21}\ J$

So, the required energy is $9.62\times 10^{-21}\ J$ .

3 0

3 years ago

Read 2 more answers

According to the Bohr model of an atom, what happens when an electron moves from the second energy level to the third energy lev

kati45 [8]

Energy is absorbed and then released to form an emission line.

When electrons absorb energy they increase there energy level. This is only temporary and the excited electron then relaxes back down to its original energy level releasing energy.

The energy is released in form of EM radiation of a specific frequency depending on the element and how many energy levels the electron relaxes.
This forms an emission line.

4 0

3 years ago

Read 2 more answers

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

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2 years ago

Which of the following is NOT true about "yield"? a. The value of the actual yield must be given in order for the percent yield

Snowcat [4.5K]

Which of the following is NOT true about "yield"? a. The value of the actual yield must be given in order for the percent yield to be calculated. b. The percent yield is the ratio of the actual yield to the theoretical yield. c. The actual yield may be different from the theoretical yield because reactions do not always go to completion. d. The actual yield may be different from the theoretical yield because insufficient limiting reagent was used.

The correct answer is:
 d. The actual yield may be different from the theoretical yield because insufficient limiting reagent was used.

8 0

2 years ago