All categories

3 years ago

As the number of protons increases but the distance between the protons and electron REMAINS THE SAME...

Chemistry

1 answer:

Aloiza [94]3 years ago

8 0

Answer:

The correct option is;

The electronegativity increases

Explanation:

The electronegativity is the measure of an atom's ability to attract a shared electron pair. The electronegativity of an atom is dependent on the atom's atomic number and the separation distance between the electrons in the valence shell and the positively charged nucleus such that an increase in the atomic number results in an increase in electronegativity and an increase in the distance between the valence electrons and the nucleus, leads to a decrease in electronegativity.

You might be interested in

Using the accepted value for the volume of 1 gram of water at the temperature of the room that you reported above, what is the a

inessss [21]

Write procedural steps that allow you to demonstrate the sun's role in the water cycle using common material - for each explain what you are modeling and how the materials you have chosen represent nature.

Using the accepted value for the volume of 1 gram of water at the temperature of the room that you reported above, what is the accepted value for the density of water

3 0

3 years ago

Which of the following possess the greatest concentration of hydroxide ions?

jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

(a) a solution of pH 3.0

First we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11$

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

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

$11=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-11}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-11}M$

(b) a solution of 0.10 M $NH_3$

As we know that 1 mole of $NH_3$ is a weak base. So, in a solution it will not dissociates completely.

So, the $OH^-$ concentration will be less than 0.10 M

(c) a solution with a pOH of 12.

We have to calculate the $OH^-$ concentration.

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

$12=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-12}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-12}M$

(d) a solution of 0.10 M NaOH

As we know that $NaOH$ is a strong base. So, it dissociates to give $Na^+$ ion and $OH^-$ ion.

So, 0.10 M of $NaOH$ in a solution dissociates to give 0.10 M of $Na^+$ ion and 0.10 M of $OH^-$ ion.

Thus, the $OH^-$ concentration is, 0.10 M

(e) a $1\times 10^{-4}M$ solution of $HNO_2$

As we know that 1 mole of $HNO_2$ in a solution dissociates to give 1 mole of $H^+$ ion and 1 mole of $NO_2^-$ ion.

So, $1\times 10^{-4}M$ of $HNO_2$ in a solution dissociates to give $1\times 10^{-4}M$ of $H^+$ ion and $1\times 10^{-4}M$ of $NO_2^-$ ion.

The concentration of $H^+$ ion is $1\times 10^{-4}M$

First we have to calculate the pH.

$pH=-\log [H^+]$

$pH=-\log (1.0\times 10^{-4})$

$pH=4$

Now we have to calculate the pOH.

$pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10$

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

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

$10=-\log [OH^-]$

$[OH^-]=1.0\times 10^{-10}M$

Thus, the $OH^-$ concentration is, $1.0\times 10^{-10}M$

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0

3 years ago

Write a double replacement precipitation reaction A precipitate forms when aqueous solutions of nickel(II) chloride and silver(I

harkovskaia [24]

Answer: The balanced chemical equation is written below.

Explanation:

Double displacement reaction is defined as the reaction in which exchange of ions takes place.

$AB+CD\rightarrow CB+AD$

Precipitation reaction is defined as the reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

When nickel (II) chloride reacts with silver (I) nitrate, it leads to the formation of white precipitate of silver chloride and an aqueous solution of nickel (II) nitrate.

The balanced chemical equation for the above reaction follows:

$NiCl_2(aq.)+2AgNO_3(aq.)\rightarrow 2AgCl(s)+Ni(NO_3)_2(aq.)$

By Stoichiometry of the reaction:

1 mole of aqueous solution of nickel (II) chloride reacts with 2 moles of aqueous solution of silver (I) nitrate to produce 2 moles of solid silver chloride and 1 moles of aqueous solution of nickel (II) nitrate

Hence, the balanced chemical equation is written above.

3 0

3 years ago

Given the following equation, N2O(g) + NO2(g) → 3 NO(g) ΔG°rxn = -23.0 kJ

Lelechka [254]

Answer:

ΔG°rxn = -69.0 kJ

Explanation:

Let's consider the following thermochemical equation.

N₂O(g) + NO₂(g) → 3 NO(g) ΔG°rxn = -23.0 kJ

Since ΔG°rxn < 0, this reaction is exergonic, that is, 23.0 kJ of energy are released. The Gibbs free energy is an extensive property, meaning that it depends on the amount of matter. Then, if we multiply the amount of matter by 3 (by multiplying the stoichiometric coefficients by 3), the ΔG°rxn will also be tripled.

3 N₂O(g) + 3 NO₂(g) → 9 NO(g) ΔG°rxn = -69.0 kJ

8 0

3 years ago

the quantity of antimony in an ore can be determined by an oxidation-reduction titration with an oxidizing agent. The ore is dis

Basile [38]

Answer:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

Explanation:

At a redox equation, one substance is being oxidized (losing electrons), and the other is being reduced (gaining electrons). In the given reaction:

BrO₃⁻(aq) + Sb³⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq)

When it's at an acidic solution, it must be ions H⁺ on the reactant, which will form water with the oxygen, so the complete reaction is:

BrO₃⁻(aq) + Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + Sb⁵⁺(aq) + H₂O(l)

As we can see, the antimony is being oxidized (go from +3 to +5), and the Bromo is being reduced. The oxidation number of brome in the reactant, knowing that the oxidation number of O is -2, is:

x + 3*(-2) = -1

x = +5

So, it's going from +5 to -1, and the half-reactions are:

BrO₃⁻(aq) + 6e⁻ → Br⁻(aq)

Sb³⁺(aq) → Sb⁵⁺(aq) + 2e⁻

The number of electrons must be the same, so the second equation must be multiplied by 3:

3Sb³⁺(aq) → 3Sb⁵⁺(aq) + 6e⁻

Thus, the equation will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + H₂O(l)

Now, we verify the amount of the elements, which must be equal on both sides. So, we multiply H₂O by 3, and H⁺ by 6, and the balanced reaction will be:

BrO₃⁻(aq) + 3Sb³⁺(aq) + 6H⁺(aq) → Br⁻(aq) + 3Sb⁵⁺(aq) + 3H₂O(l)

6 0

3 years ago