CHEMISTRY HIGH SCHOOL

Which substance can be decomposed by chemical means?(1) ammonia (3) phosphorus
(2) oxygen (4) silicon

Answers

Answer 1

Answer:

$\boxed{\left( 1 \right){\text{ ammonia}}}$ is the substance that can be decomposed by chemical means.

Further Explanation:

The pure form of matter is defined as substance but the mixture is termed as the combination of atoms and molecules.

Classification of substances:

(a). Element

It is a type of pure substance and is the simplest form that cannot be broken down by any chemical means. Copper, iron, and aluminium are some of the examples of elements.

(b) Compound

It is composed of two or more different elements that are held together by chemical bonds. These can be decomposed into their respective constituents. Compounds have no similarity in properties with those of their constituent particles. NaCl, ${\text{C}}{{\text{O}}_{\text{2}}}$ and ${\text{N}}{{\text{H}}_3}$ are examples of compounds.

(1) Ammonia contains one nitrogen and three hydrogen atoms so it is a compound. Since compounds can be decomposed chemically, ammonia is also decomposed by chemical means.

(2) Oxygen (O) is an element so it is the simplest form in which it can exist. Therefore it cannot be decomposed by chemical means.

(3) Phosphorus (P) is an element so it is the simplest form in which it can exist. Therefore it cannot be decomposed by chemical means.

(4) Silicon (Si) is an element so it is the simplest form in which it can exist. Therefore it cannot be decomposed by chemical means.

Learn more:

Give an example of a two-phase mixture: brainly.com/question/838401
Which sample is a pure substance? brainly.com/question/2227438

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Elements, compounds, and mixtures

Keywords: substance, ammonia, oxygen, phosphorus, silicon, element, compound, chemical means, decomposed, broken down, simplest form.

Answer 2

Answer:

Final answer:

Ammonia is a compound composed of nitrogen and hydrogen, and as such, can be chemically decomposed into these elements. Oxygen, phosphorus and silicon, being elements, cannot be decomposed by chemical means.

Explanation:

In the context of chemistry, a substance can be decomposed by chemical means if it is a compound. Compounds are composed of two or more different elements and can be broken down into those elements through chemical reactions. In the given options, (1) ammonia is a compound, composed of nitrogen and hydrogen elements. Therefore, ammonia can be chemically decomposed into its constituent elements. In contrast, oxygen, phosphorus, and silicon are all elements, which means they cannot be decomposed further by chemical means.

Learn more about Decomposition of Substances here:

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Related Questions

State charles's law. If a balloon occupies 885 cm3 volume at 20°C and 794 cm3 at - 10°C , prove that this is according to charles's law.

Answers

find the distance between

the following points:

a) AC0₂0) and B(5, 2)

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter? NaOH (s) → Na+ (aq) + OH– (aq) ∆H = -44.5 kJ/mol

Answers

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=(q)/(n)$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=(1.52g)/(40g/mole)=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=(q)/(0.038mol)$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m* c* (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g* 4.18J/g^oC* (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

Q(0,5) and R(2,1) are the endpoints of a line segment. What is the midpoint M of that line segment.

Answers

To find the midpoint M of a line segment with endpoints Q(0,5) and R(2,1), you can use the midpoint formula:

Midpoint M = ((x₁ + x₂) / 2, (y₁ + y₂) / 2)

In this formula:

- (x₁, y₁) are the coordinates of the first endpoint (Q in this case).

- (x₂, y₂) are the coordinates of the second endpoint (R in this case).

Plug in the values:

M = ((0 + 2) / 2, (5 + 1) / 2)

M = (2 / 2, 6 / 2)

M = (1, 3)

So, the midpoint M of the line segment with endpoints Q(0,5) and R(2,1) is (1, 3).

As a Ca atom undergoes oxidation to Ca2+, the number of neutrons in its nucleus(1) decreases
(2) increases
(3) remains the same

Answers

Answer:

Hence the correct answer is remains the same

Explanation:

As a Ca atom undergoes oxidation to Ca2+,as follows:

Ca --- Ca 2+ + 2e-

In the process of oxidation calcium loss 2 electron only there is no change in the number of protons as well neutrons therefore there is no change in the number of neutrons in its nucleus

Hence the correct answer is remains the same

Oxidation:

Oxidation is a process in which either 1 or all following changes occurs:

1. Gaining of oxygen atoms

2. Loss of electrons

3. Loss of hydrogen atom.

4. Increasing oxidation number.

Reduction:

Reduction is a process in which either 1 or all following changes occurs:

1. Loss of oxygen atoms

2. Gaining of electrons

3. Gaining of hydrogen atom.

4. Decreasing oxidation number.

Oxidation refers to a change in electrons (In this case, loss of electrons). It has no impact on the number of neutrons, or for that matter on any sub-atomic particles in the nucleus. So the number of neutrons (3) remains the same

The radioisotope radon-222 has half-life of 3.8 days. how much of a 73.9-g sample of radon-222 would be left after approximately 23 days?

Answers

This can be solved by the equation:

A = Aoe^(-ln2/t)T
where Ao is the initial amount of the substance ( 73.9-g sample of radon-222 ), t is the half-life time (3.8 days) and T is the time elapsed. Plugging in the values in the equation, the substance left is 1.11g.

How does latitude affect a climate zone?(A) Climate zones at low latitudes are hotter because they receive more direct sunlight.

(B) Climate zones at high latitudes are cooler because they are closer to the equator.

(C) Climate zones at high latitudes receive a lot of precipitation because they are farther from the equator.

(D) Climate zones at low latitudes receive little precipitation because they are closer to the ocean.

Answers

(A) Climate zones at low latitudes are hotterbecause they receive more direct sunlight. This is because the Earth's surface at the equator is closer to the sun and therefore receives more direct sunlight

Additionally, the distance of a region from the equator affects its precipitation pattern, with regions closer to the equator generally receiving more precipitation than regions at higher latitudes.

As a result, the tropics, which lie near the equator, tend to be warmer than other regions. In contrast, regions at high latitudes receive less direct sunlight due to the curvature of the Earth's surface, resulting in colder temperatures.

To know more about climate zone, click here:-

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