A student needs .575 moles of sodium chloride for an experiment. how many grams she she mass out

The electron is the primary atomic particle that determines an atom's chemical behaviour.

Electrons are negatively charged particles that form electron shells or energy levels around an atom's nucleus. They are engaged in the production and breaking of chemical bonds, hence they play an important part in chemical reactions.

The quantity and configuration of electrons in an atom's outermost energy level (valence electrons) are very significant in determining the chemical characteristics of the atom. Because of the identical number of valence electrons, elements in the same group or column of the periodic table frequently show similar chemical behaviour.

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Answer: Option (a) is the correct answer.

Explanation:

A reduction-half reaction is defined as the reaction in which there occurs gain of one or more number of electrons by a specie or element.

Or, in a reduction-half reaction there occurs loss of oxygen if it is present in a reaction.

But in a reduction-half reaction loss of electrons will always take place.

For example,

Reduction-half reaction:

So here, oxidation state of copper is changing from +2 to 0. Also, loss of oxygen is occurring in the reaction equation.

Oxidation-half reaction:

Thus, we can conclude that in a reduction half-reaction elements gain electrons.

Final answer:

In a calcium atom, 3) an electron in the third shell has more energy than an electron in the second shell. This is because the shell's distance from the nucleus determines the energy of an electron: the farther the shell, the higher the energy.

Explanation:

The relative energy of the electrons in the shells of a calcium atom would be best described by statement 3, which states: 'An electron in the third shell has more energy than an electron in the second shell.' This is due to the fact that electrons occupy shells (also known as energy levels) around an atom's nucleus, and the further an electron is from the nucleus, the higher its energy. This is because it is closer to the outer environment and has a higher potential energy due to its distance from the atom’s positive core.

In the case of calcium, which has an atomic number of 20, it has four energy levels. This means that an electron in the fourth shell would certainly have more energy than an electron in the first, second, or third shells, and an electron in the third shell would have more energy than one in the second or first shells.

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Answer:

-909.3KJ/mole

Explanation:

The heat of reaction is accessible from the heat of formation of reactants and products using the formula below:

ΔH = Σ ΔHf products - Σ ΔHf reactants

Before we proceed, it is important to know that the enthalpy of formation of element is zero ,be it a single element or a molecule of an element.

From the reaction for the formation of sulphuric acid, we know we need to know the heat of formation of sulphur (vi) oxide and water. The examiner is quite generous and have us for water already.

Now we need to calculate for sulphur (vi) oxide. This is calculated as follows:

We first calculate for sulphur(iv)oxide. This can be obtained from the reaction between sulphur and oxygen. The calculation goes thus:

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole suphur(iv) oxide × x] - [ (1 mole of elemental sulphur × 0) + (1 mole of elemental oxygen × 0]

We were already told this is equal to -296.8KJ. Hence the heat of formation of sulphur(iv) oxide is -296.8KJ.

We then proceed to the second stage.

Now, here we have 1 mole sulphur (iv) oxide reacting with 0.5 mole oxygen molecule.

We go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphur (vi) oxide × y] - [ (1 mole of sulphur (iv) oxide × -296.8) + (0.5 mole of oxygen × 0)].

We already know that the ΔH here equals -98.9KJ.

Hence, -98.9 = y + 296.8

y = -296.8KJ - 98.9KJ = -395.7KJ

We now proceed to the final part of the calculation which ironically comes first in the series of sentences.

Now, we want to calculate the standard heat of formation for sulphuric acid. From the reaction, we can see that one mole of sulphur (vi) oxide, reacted with one mole of water to yield one mole of sulphuric acid.

Mathematically, we go again :

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔH = [ 1 mole of sulphuric acid × z] - [( 1 mole of sulphur vi oxide × -395.7) + ( 1 mole of water × -285.8)].

Now, we know that the ΔH for this particular reaction is -227.8KJ

We then proceed to to open the bracket.

-227.8 = z - (-395.7 - 285.8)

-227.8 = z - ( -681.5)

-227.8 = z + 681.5

z = -227.8-681.5 = -909.3KJ

Hence, ΔH∘f for sulphuric acid is -909.3KJ/mol

Final answer:

The total number of neutrons in an atom of O-18 is 10. This is determined by subtracting the atomic number (number of protons) from the atomic mass (number of protons + neutrons).

Explanation:

To find the number of neutrons in an atom, you subtract the atomic number (number of protons) from the atomic mass (number of protons + neutrons). In an atom of O-18 (Oxygen-18), the atomic number of oxygen is 8, which means there are 8 protons. The 18 in O-18 represents the atomic mass. Subtracting the atomic number (8) from the atomic mass (18) gives us the number of neutrons, which is 10.

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