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3 years ago

PLEASE HELP ASAP Which one of the following is a step used for balancing chemical equations?

Physics

1 answer:

scZoUnD [109]3 years ago

3 0

Answer:

C. Count the atoms in each substance in the reactants and products.

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

In order for a chemical equation to be balanced, the condition which must be met is that the number of atoms in the reactants equals the number of atoms in the products.

This ultimately implies that, the mass and charge of the chemical equation are both balanced properly.

In Chemistry, all chemical equation must follow or be in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed by either a physical transformation or a chemical reaction but transformed from one form to another in an isolated (closed) system.

One of the step used for balancing chemical equations is to count the atoms in each substance in the reactants and products.

For example;

NH3 + O2 -----> NO + H2O

The number of atoms in each chemical element are;

For the reactant side:

Nitrogen, N = 1

Hydrogen, H = 3

Oxygen, O = 2

For the product side;

Nitrogen, N = 1

Hydrogen, H = 2

Oxygen, O = 2

When we balance the chemical equation, we would have;

NH3 + 3O2 -----> 4NO + 2H2O

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andrey2020 [161]

Assuming air as ideal gas and amount of air in no of moles is known then by gas law,

PV= nRT

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4 years ago

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

E = 0

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u /> $E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$ <u />

<u /> $E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$ <u />

Again, the minus sign indicates the direction of the field towards the center.

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

4 years ago

What is the maximum number of primary partitions that gpt supports?

Vikki [24]

It supports 128 primary partitions.

4 0

4 years ago

"In a pure substance all the particles must be identical; therefore pure substances are composed only of elements." Do you agree

kenny6666 [7]

I do not agree with the statement.
The "substance" can be a compound. It's "pure"
as long as there's nothing else in it but its name.

'Pure' water is 100% H₂O with nothing else in it.
'Pure' table salt is 100% NaCl with nothing else in it.
'Pure' carbon dioxide is 100% CO₂ with nothing else in it.

These example substances are all compounds, not elements.

6 0

3 years ago

2 a A pile of 60 sheets of paper is 6 mm high. Calculate the average thickness of a sheet of the paper.

fiasKO [112]

The average thickness of a sheet of the paper is 0.1 mm.

The number of ice blocks that can be stored in the freezer is 80 blocks of ice.

<h3>Average thickness of a sheet of the paper</h3>

The average thickness of a sheet of the paper is calculated as follows;

average thickness = 6 mm/60 sheets = 0.1 mm /sheet

Thus, the average thickness of a sheet of the paper is 0.1 mm.

<h3>Volume of each block of ice</h3>

Volume = 10 cm x 10 cm x 4 cm

Volume = 400 cm³

<h3>Volume of the freezer</h3>

Volume = 40 cm x 40 cm x 20 cm = 32,000 cm³

<h3>Number of ice blocks that can be stored</h3>

n = 32,000 cm³/400 cm³

n = 80 blocks of ice

Thus, the number of ice blocks that can be stored in the freezer is 80 blocks of ice.

Learn more about average thickness here: brainly.com/question/24268651

#SPJ1

6 0

1 year ago