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deff fn [24]
3 years ago
5

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

Chemistry
1 answer:
Alchen [17]3 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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What is the percent of copper in CuS?
bonufazy [111]

Answer:

Element Symbol Mass Percent

Cuprum Cu 66.464%

Sulfur S 33.537%

Explanation:

I got this out of my module, sorry if it's wrong but i am pretty sure 97% this is correct!

7 0
3 years ago
Explain why beryllium is produced when potassium is heated with beryllium
Vladimir [108]

Explanation:

The more reactive element replaces less reactive element during chemical reaction.

Since, potassium is more reactive than beryllium. When potassium reacts with beryllium choride, it replaces beryllium and forms potassium chloride and produces beryllium.

3 0
3 years ago
If 2.60 g of NaBr are dissolved in enough water to make 160. mL of solution, what is the molar concentration of
navik [9.2K]

This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.

<h3>Molarity</h3>

In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

M=\frac{n}{V}

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

2.60g*\frac{1mol}{102.89g} =0.0253mol

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

M=\frac{0.0253mol}{0.160L}=0.158M

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

V=\frac{n}{M} =\frac{0.158M}{0.120mol/L}\\ \\V=0.211L

That in milliliters turns out to be:

V=0.211L*\frac{1000mL}{1L}=211mL

Learn more about molarity: brainly.com/question/10053901

6 0
2 years ago
I really need help..please and thank you.
Elenna [48]

What is the question you need answered

8 0
3 years ago
KClO3 ---&gt; KCl + O2 Balance the decomposition chemical reaction. A) KClO3 ---&gt; KCl + O2 B) 2KClO3 ---&gt; KCl + 3O2 C) 2KC
Damm [24]
And the answer is C.
5 0
3 years ago
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