What do atoms when they share electron pairs

What is true when a reaction has reached equilibrium?

jolli1 [7]

Answer:

Explanation:

The rate of the forward reaction is less than the rate of the reverse reaction. ... The reaction rates of the forward and reverse reactions are equal

4 0

3 years ago

What mass of potassium hypochlorite (FW-90.6 g/mol) must be added to 4.50 x 10 mL of water to give a solution with pH 10.20? [Ka

marta [7]

Answer : The mass of potassium hypochlorite is, 4.1 grams.

Explanation : Given,

pH = 10.20

Volume of water = $4.50\times 10^2ml=0.45L$

The decomposition of KClO will be :

$KClO\rightarrow K^++ClO^-$

Now the further reaction with water $(H_2O)$ to give,

$ClO^-+H_2O\rightarrow HClO+OH^-$

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-10.20=3.8$

Now we have to calculate the $OH^-$ concentration.

$pOH=-\log [OH^-]$

$3.8=-\log [OH^-]$

$[OH^-]=1.58\times 10^{-4}M$

Now we have to calculate the base dissociation constant.

Formula used : $K_b=\frac{K_w}{K_a}$

Now put all the given values in this formula, we get :

$K_b=\frac{1.0\times 10^{-14}}{4.0\times 10^{-8}}=2.5\times 10^{-7}$

Now we have to calculate the concentration of $ClO^-$ .

The equilibrium constant expression of the reaction is:

$K_b=\frac{[OH^-][HClO]}{[ClO^-]}$

As we know that, $[OH^-]=[HClO]=1.58\times 10^{-4}M$

$2.5\times 10^{-7}=\frac{(1.58\times 10^{-4})^2}{[ClO^-]}$

$[ClO^-]=0.0999M$

Now we have to calculate the moles of $ClO^-$ .

$\text{Moles of }ClO^-=\text{Molarity of }ClO^-\times \text{Volume of solution}$

$\text{Moles of }ClO^-=0.0999mole/L\times 0.45L=0.0449mole$

As we know that, the number of moles of $ClO^-$ are equal to the number of moles of KClO.

So, the number of moles of KClO = 0.0449 mole

Now we have to calculate the mass of KClO.

$\text{Mass of }KClO=\text{Moles of }KClO\times \text{Molar mass of }KClO$

$\text{Mass of }KClO=0.0449mole\times 90.6g/mole=4.07g\approx 4.1g$

Therefore, the mass of potassium hypochlorite is, 4.1 grams.

6 0

3 years ago

Light from three different lasers (A, B, and C), each with a different wavelength, was shined on the same metal surface. Laser A

kenny6666 [7]

Explanation:

It is known that each metal has a different work function and it is just the energy required to remove the electron from the surface.

The relation between energy and work function is as follows.

$\Delta E = Q + K.E$

where, $\Delta E$ = energy of photon

Q = work function

K.E = kinetic energy of electron

When the value of Q is large then the amount of energy required by the photon is also large and vice-versa.

And, when there is not sufficient energy to remove an electron from metal surface or energy is less than its work function then electron will not be removed.

On the other hand, when incoming photon will have enough energy then electron will be ejected.

In the given situation, light from laser A is not able to remove the electron from metal surface which means that energy of photon is less than work function Q.

As, both B and C are able to remove electron from the surface of metal. So, after the removal of electrons rest of the energy of photons is transferred to eject electrons and this energy is converted into kinetic energy.

As energy of B is greater than the energy of C and A has the lowest energy among all.

Therefore, order of energy from lowest to highest is as follows.

A < C < B

Now, the relation between energy and wavelength is as follows.

E = $\frac{hc}{\lambda}$

where, $\lambda$ = wavelength

So, energy is inversely proportional to wavelength. Therefore, increase in energy means decrease in wavelength.

Therefore, increasing order of wavelength is as follows.

B < C < A

8 0

3 years ago

How do electrons in a metallic bond behave?

igomit [66]

Hello,

In metallic bonds, the valence electrons from the s and p orbitals of the interacting metal atoms delocalize. That is to say, instead of orbiting their respective metal atoms, they form a “sea” of electrons that surrounds the positively charged atomic nuclei of the interacting metal ions.

Hope this helps!!!! Happy Holidays!!!! (:

4 0

3 years ago

Consider the following equilibrium systems: A⇌2B ΔH o =20.0 kJ/mol Reaction 1 A+B⇌C ΔH o =−5.4 kJ/mol Reaction 2 A⇌B ΔH o =0.0 k

galina1969 [7]

Answer:

Reaction 1: Kc increases

Reaction 2: Kc decreases

Reaction 3: The is no change

Explanation:

Let us consider the following reactions:

Reaction 1: A ⇌ 2B ΔH° = 20.0 kJ/mol

Reaction 2: A + B ⇌ C ΔH° = −5.4 kJ/mol

Reaction 3: 2A⇌ B ΔH° = 0.0 kJ/mol

To predict what will happen when the temperature is raised we need to take into account Le Chatelier Principle: when a system at equilibrium suffers a perturbation, it will shift its equilibrium to counteract such perturbation. This means that <em>if the temperature is raised (perturbation), the system will react to lower the temperature.</em>

Reaction 1 is endothermic (ΔH° > 0). If the temperature is raised the system will favor the forward reaction to absorb heat and lower the temperature, thus increasing the value of Kc.

Reaction 2 is exothermic (ΔH° < 0). If the temperature is raised the system will favor the reverse reaction to absorb heat and lower the temperature, thus decreasing the value of Kc.

Reaction 3 is not endothermic nor exothermic (ΔH° = 0) so an increase in the temperature will have no effect on the equilibrium.

8 0

2 years ago