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3 years ago

The smallest unit of an element that retains all the characteristics of that element is called a(n)

1 answer:

3 0

<span>The smallest unit of an element that retains all the characteristics of that element is called an A. atom.
Atoms consist of protons, electrons, and neutrons, but they are particles and don't have the characteristics of the element which is why D is incorrect. B and C are not the smallest units - atoms are smaller than them.
</span>

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A) explain why caesium is more reactive than potassium.

Kamila [148]

Answer:

Explanation:

(a) Firstly, caesium abd potassium are both in Group 1 of the periodic table. Group 1 metals (also called alkali metals) are the most reactive metals of the periodic table. Caesium is more reactive than Potassium because it has a higher electropositivity than Potassium. Electropositivity is the tendency of a metal to donate electron(s) to form a cation. Electropositivity increases down the group; this is because it is easier for atoms to loose electrons on the outermost shell that are far away from the central nucleus as against atoms whose outermost electrons are closer to the central nucleus. <u>Thus, the more "bulky" an atom is, the farther it's outermost electrons (valence electrons) get from the central nucleus and the easier it is to lose the outermost electron(s). And the easier it is for the valence electron(s) to be removed, the more reactive the atom would be and vice-versa.</u>

Caesium is more reactive than potassium because it is more bulky than potassium, with an atomic number of 55, while potassium has an atomic number of 19.

NOTE: The closer an electron is to the nucleus, the more difficult it is to be removed from it's shell.

(b) i. Formula for Caesium Nitrate:

Symbol for Caesium is Cs and Nitrate is NO₃⁻.

Cs⁺ + NO₃⁻ ↔ CsNO₃

Formula for Caesium Nitrate is CsNO₃

ii. Formula for Caesium sulphate

Symbol for caesium is Cs and Sulphate is SO₄²⁻

Cs⁺ + SO₄²⁻ ↔ Cs₂SO₄

Formula for Caesium sulphate is Cs₂SO₄

NOTE: When writing the formulae, the charges would be exchanged to form the subscript as seen on the product sides above.

7 0

3 years ago

The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?

Vedmedyk [2.9K]

Answer : The volume of pure diamond is $0.493inch^3$

Explanation : Given,

Density of pure carbon in diamond = $3.52g/cm^3$

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

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

Molar mass of carbon = 12 g/mol

$\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g$

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

$Density=\frac{Mass}{Volume}$

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

$3.52g/cm^3=\frac{284.4g}{Volume}$

Volume = $80.8cm^3$

As we know that:

$1cm^3=0.061inch^3$

So,

Volume = $0.061\times 80.8inch^3$

Volume = $0.493inch^3$

Therefore, the volume of pure diamond is $0.493inch^3$

5 0

3 years ago

Matter can be measured in many ways for specific purposes which measurement depends on the force of gravity and is not a reliabl

dlinn [17]

Answer:True But matter can take up much space and is stronger than density and gravity

Explanation: Gravity is like a force that just pull on objects now if you get down to the magnets well magnetism is the strongest force on earth that could be connected through the core

7 0

3 years ago

Read 2 more answers

Which part of the wave does the arrow point to?

AlekseyPX

I think it’s gonna be D

7 0

3 years ago

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

3 years ago