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deff fn [24]
2 years ago
5

How can energy transfer affect the physical and chemical properties of matter?

Chemistry
1 answer:
Alchen [17]2 years ago
7 0

A physical change is a change in the state or properties of matter without any accompanying change in the chemical identities of the substances contained in the matter. Physical changes are observed when wax melts, when sugar dissolves in coffee, and when steam condenses into liquid water. Other examples of physical changes include magnetizing and demagnetizing metals (as is done with common anti-theft security tags) and grinding solids into powders (which can sometimes yield noticeable changes in color).

The characteristics that distinguish one substance from another are called properties. A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. Some physical properties, such as density and color, may be observed without changing the physical state of the matter. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes a physical change

Figure :Copper and nitric acid undergo a chemical change to form copper nitrate and brown, gaseous nitrogen dioxide. During the combustion of a match, cellulose in the match and oxygen from the air undergo a chemical change to form carbon dioxide and water vapor. Cooking red meat causes a number of chemical changes, including the oxidation of iron in myoglobin that results in the familiar red-to-brown color change.

chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO2 and H2O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

(1 CO2 molecule × 2 O atoms per CO2 molecule) + (2 H2O molecules × 1 O atom per H2O(molecule) = 4 O atoms

The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here: CH 4 + 2O2⟶CO2 + 2H2O

Element Reactants Products Balanced?

C 1×1 = 1 1×1 = 1 1 = 1, yes

H 4×1 = 4 2×2 = 4 4 = 4, yes

O 2×2 = 4 (1×2) + (2×1) = 4 4 = 4, yes

A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an unbalanced chemical equation: H2O⟶H2 + O2 (unbalanced)

Comparing the number of H and O atoms on either side of this equation confirms its imbalance:

Element Reactants Products Balanced?

H 1×2 = 2 1×2 = 2 2 = 2, yes

O 1×1 = 1 1×2 = 2 1 ≠ 2, no

H2O to H2 O2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H2O to 2.

2H2O⟶H2 + O2 (unbalanced)

Element Reactants Products Balanced?

H 2×2 = 4 1×2 = 2 4 ≠ 2, no

O 2×1 = 2 1×2 = 2 2 = 2, yes

The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H2 product to 2.

2H2O⟶2H2 + O2 (balanced)

Element Reactants Products Balanced?

H 2×2 = 4 2×2 = 4 4 = 4, yes

O 2×1 = 2 1×2 = 2 2 = 2, yes

These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:

2H2O⟶2H2 + O2

Types of Chemical Reactions

Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they hit each other with their fists, we say they are fighting

A photograph is shown of a yellow green opaque substance swirled through a clear, colorless liquid in a test tube.

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The kinetic energy of the emitted electrons of cesium when it is exposed to UV rays of frequency 1.0 \times {10^{15}}\;{\text{Hz}}  is  \boxed{6.63 \times {{10}^{ - 19}}\;{\text{J}}}

Further Explanation:

Photoelectric effect:

When light is made to fall on any substance, electrons are emitted from it. This is known as the photoelectric effect and the emitted electrons are called photoelectrons. The electrons are emitted because of the transference of energy from light to the electrons.

Cesium is a member of the alkali metal group so it is highly reactive and shows photoelectric effect to the maximum extent. It can remove its electron so easily because of its atomic size. Due to large atomic size of cesium, its outermost electrons are held very less tightly to the nucleus and therefore removed easily.

According to the Planck-Einstein equation, the energy is proportional to the frequency and is expressed as follows:

{\mathbf{E=h\nu }}                                   ......(1)

Here,

{\text{E}} is the energy.

h is the Plank’s constant.

\nu is the frequency.

The frequency of UV rays is 1.0 \times {10^{15}}\;{\text{Hz}} or 1.0 \times {10^{15}}\;{{\text{s}}^{ - 1}}

The value of Planck’s constant is 6.626 \times {10^{ - 34}}\;{\text{J}}\cdot{\text{s}} .

Substitute these values in equation (1)

\begin{aligned}{\text{E}}&=\left( {6.626 \times {{10}^{ - 34}}\;{\text{J}}\cdot{\text{s}}}\right)\left( {1.0 \times {{10}^{15}}\;{{\text{s}}^{ - 1}}}\right)\\&=6.63\times {10^{ - 19}}\;{\text{J}}\\\end{aligned}

But when electrons are ejected out from the surface of the substance, all of its energy is considered as kinetic energy.

So the kinetic energy of the electrons is {\mathbf{6}}{\mathbf{.63 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 19}}}}\;{\mathbf{J}} .

Learn more:

1. Statement about subatomic particle: brainly.com/question/3176193

2. The energy of a photon in light: brainly.com/question/7590814

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of the atom

Keywords: kinetic energy, frequency, energy, photoelectric effect, Planck's constant, light, electrons, photoelectrons, proportional, transference, reactive, cesium.

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