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3 years ago

In order for a nonmetal to obey the octet rule, it____.

Chemistry

2 answers:

lyudmila [28]3 years ago

6 0

Answer: Option (C) is the correct answer.

Explanation:

Non-metals are the species which have deficiency of electrons. Therefore, in order to gain stability non-metals tend to gain electrons so that their octet gets completely filled.

For example, atomic number of chlorine is 17 and its electrons are distributed as 2, 8, 7. Thus, in order to completely filled the shell chlorine will gain one electron from an electron rich atom or metal.

Hence, we can conclude that in order for a nonmetal to obey the octet rule, it gains electrons.

lubasha [3.4K]3 years ago

3 0

C. In order for a nonmetal to obey the octet rule, it gains electrons.

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PLEASE HELP!!! PLEASE.

adell [148]

Answer:

Q₁: [HCl] = 0.075 N = 0.075 M.

Q₂: [KOH] = 7.675 mN = 7.675 mM.

Q₃: [H₂SO₄] = 0.2115 N = 0.105 M.

Q₄: The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

Explanation:

Q₁: If it takes 67 mL of 0.15 M NaOH to neutralize 134 mL of an HCl solution, what is the concentration of the HCl?

As acid neutralizes the base, the no. of gram equivalent of the acid is equal to that of the base.

The normality of the NaOH and HCl = Their molarity.

∵ (NV)NaOH = (NV)HCl

∴ N of HCl = (NV)NaOH / (V)HCl = (0.15 N)(67 mL) / (134 mL) = 0.075 N.

∴ The concentration of HCl = 0.075 N = 0.075 M.

Q₂: If it takes 27.4 mL of 0.050 M H₂SO₄ to neutralize 357 mL of KOH solution, what is the concentration of the KOH solution?

As mentioned in Q1, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

The normality of H₂SO₄ = Molarity of H₂SO₄ x 2 = 0.050 M x 2 = 0.1 N.

∵ (NV)H₂SO₄ = (NV)KOH

∴ N of KOH = (NV)H₂SO₄ / (V)KOH = (0.1 N)(27.4 mL) / (357 mL) = 7.675 x 10⁻³ N = 7.675 mN.

∴ The concentration of KOH = 7.675 mN = 7.675 mM.

Q₃:If it takes 55 mL of 0.5 M NaOH solution to completely neutralize 130 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?

As mentioned in Q1 and 2, the no. of gram equivalent of the acid is equal to that of the base at neutralization.

The normality of NaOH = Molarity of NaOH = 0.5 N.

∵ (NV)H₂SO₄ = (NV)NaOH

∴ N of H₂SO₄ = (NV)NaOH / (V)H₂SO₄ = (0.5 N)(55 mL) / (130 mL) = 0.2115 N.

∴ The concentration of H₂SO₄ = 0.2115 N = 0.105 M.

Q₄: Explain the difference between an endpoint and equivalence point in a titration.

The equivalence point is the point at which the added titrant is chemically equivalent completely to the analyte in the sample whereas the endpoint is the point where the indicator changes its color.

The equivalence point in a titration is the point at which the added titrant is chemically equivalent completely to the analyte in the sample. It comes before the end point. At the equivalence point, the millimoles of acid are chemically equivalent to the millimoles of base.

End point is the point where the indicator changes its color. It is the point of completion of the reaction between two solutions.

The effectiveness of the titration is measure by the close matching between equivalent point and the end point. pH of the indicator should match the pH at the equivalence to get the same equivalent point as the end point.

6 0

3 years ago

Everyone needs a certain number of ____________ to meet their daily energy needs.

Ganezh [65]

Answer:

The answer is calories.

The average human being needs 2000 calories for them to be energized.

8 0

2 years ago

Which law is associated with inertia

Doss [256]

Newton's first law is the answer.

5 0

3 years ago

Compared to the charge of a proton, the charge of all electron has

Kay [80]

Answer: option (4) the same magnitude and the opposite sign.

Justification:

1) Electrons are negative particles thar are around the nucleus of the atom (in regions called orbitals).


2) Protons are positive particles that are inside the nuclus of the atom.


3) The nucleus of the atom has the same number of protons as electrons are in the orbitals of the atom.


4) The atoms are neutral (neither positive nor negative) because there are the same number of electrons and protons and their charge are of the same magnitude but different sign: (+) + (-) = 0: positive + negative = neutral.

6 0

2 years ago

Read 2 more answers

Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)

saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if $\rm O_2$ is removed from the mixture. The reason is that collisions between $\rm SO_2$ molecules and $\rm O_2\!$ molecules would become less frequent.

The reaction would not be at equilibrium for a while after $\rm O_2$ was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of $\rm SO_3\, (g)$ to the system, the backward reaction would have broken down the same amount of $\rm SO_3\, (g)\!$ . So is the case for $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ .

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield $\rm SO_3\, (g)$ , only a fraction of the reactions will be fruitful.

Assume that $\rm O_2\, (g)$ molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer $\!\rm O_2\, (g)$ molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between $\rm SO_2\, (g)$ and $\rm O_2\, (g)\!$ molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after $\rm O_2\, (g)$ was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more $\rm SO_3\, (g)$ gets converted to $\rm SO_2\, (g)$ and $\rm O_2\, (g)$ , the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0

2 years ago