Atmospheric pollution is the worst when: natural gas is burned coal is burned geothermal heat is used to generate electricity

An exciton is: A charged particle made of electrons A charged particle made of holes A neutral particle made of an electron and

Alex

Answer:

A neutral particle made of an electron and hole

Explanation:

Exciton

It is the combination of an electron and a hole ( hole refers to the vacancy of an electron ) . And , as both the electron and the hole have the same charge but the polarity is opposite , the combination will lead to a neutral compound , i.e. , Exciton have no charge and so neutral .

It is free to move in the nonmetallic crystal and since it charge less , it is difficult to detect it directly .

7 0

3 years ago

How many moles of HNO3 are present if 4.90×10−2 mol of Ba(OH)2 was needed to neutralize the acid solution?

Amiraneli [1.4K]

Answer:

0.098 moles

Explanation:

Let y represent the number of moles present

1 mole of Ba(OH)₂ contains 2 moles of OH- ions.

Hence, 0.049 moles of Ba(OH)2 contains y moles of OH- ions.

To get the y moles, we then do cross multiplication

1 mole * y mole = 2 moles * 0.049 mole

y mole = 2 * 0.049 / 1

y mole = 0.098 moles of OH- ions.

1 mole of OH- can neutralize 1 mole of H+

Therefore, 0.098 moles of HNO₃ are present.

3 0

3 years ago

Consider the reaction: P(s) + 5/2 Cl2(g)PCl5(g) Write the equilibrium constant for this reaction in terms of the equilibrium con

Pani-rosa [81]

Answer: The equilibrium constant for the overall reaction is $K_a\times K_b$

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) $P(s)+\frac{3}{2}Cl_2(g)\rightarrow PCl_3(g)$

$K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}$

b) $PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)$

$K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}$

For overall reaction on adding a and b we get c

c) $P(s)+\frac{5}{2}Cl_2(g)\rightarrow PCl_5(g)$

$K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}$

$K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}$

The equilibrium constant for the overall reaction is $K_a\times K_b$

4 0

3 years ago

Palladium (Pd; Z 46) is diamagnetic. Draw partial orbital diagrams to show which of the following electron configurations is con

nexus9112 [7]

Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom are determined by the electronic configuration.

Paramagnetic compounds : They have unpaired electrons.

Diamagnetic compounds : They have no unpaired electrons that means all are paired.

The given electron configurations of Palladium are:

(a) [Kr] 5s²4d⁸

In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.

(b) [Kr] 4d¹⁰

In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.

(c) [Kr] 5s¹4d⁹

In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.

6 0

3 years ago

According to bohr's model of the atom in which orbitals do electrons have the least energy

sasho [114]

According to Bohr's model of the atom, the higher the orbital in which the electrons are found, the higher their energy or excitation state. Therefore, the electrons with the least amount of energy are those at the lowest orbitals, which are closer to the nucleus.
These orbitals are characterized by 4 quantum numbers, namely the principal quantum number (n), orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms). The principal quantum number reflects the distance of the electrons from the nucleus with n=1 as the orbital closest to the nucleus. Thus, according to Bohr's model, electrons in the orbital with n=1 have the lowest energy.

7 0

3 years ago