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bija089 [108]
3 years ago
9

What is the difference between physical and chemical properties? Give 3 examples of each.

Chemistry
1 answer:
IceJOKER [234]3 years ago
7 0

Explanation:

1.A physical property is an aspect of matter that can be seen or measured without changing its chemical composition. Examples of physical properties include color, molecular weight, and volume.

A chemical property is observed only by changing the chemical identity of a substance. In other words, the only way to detect a chemical is to perform a chemical reaction.

2.This property measures the ability of chemical change. Examples of chemical properties are reactivity, flammability, and oxidation state.

The physical properties of a substance do not involve any chemical reaction. These include density, color, mass, hardness, freezing points, electrical properties, and the like.

Chemical properties include the reaction of chemicals with other substances. These reactions lead to the disappearance of the raw material and the appearance of new materials that have different physical and chemical properties.

3.Chemical properties can be compared to physical properties; On the contrary, they are recognizable without changing the structure of matter. However, for many properties in the field of physical chemistry and other disciplines at the boundary between chemistry and physics, the distinction can be a matter for the researcher's point of view. The properties of materials, both physical and chemical, can be seen as metaphysical; This means that it is secondary to the principle of tangible reality. Multiple metamorphic layers are also possible.

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A 60g sample of tetraethyl lead, a gasoline additive, is found to contain 38.43g lead, 17.83g carbon and 3.74g hydrogen. Determi
anastassius [24]

Answer:

Empirical formula (which matches the molecular formula) is = PbC₈H₂₀

Explanation:

Our sample: 60 g of tetraethyl lead

In order to determine the compound empirical formula we need the centesimal composition:

(Mass of element / Total mass) . 100 =

(38.43 g lead / 60g ) . 100 = 64.05%

(17.83 g C / 60g) . 100 = 29.72%

(3.74 g H / 60g) . 100 = 6.23 %

These % are the mass of the elements in 100 g of compound. Let's find out the moles of them:

64.05 g / 207.2 g/mol = 0.309 moles

29.72 g / 12 g/mol = 2.48 moles

6.23 g/ 1 g/mol = 6.23 moles

Next, we divide the moles, by the lowest value of them (0.309)

0.309 / 0.309 = 1 mol Pb

2.48 / 0.309 = 8 mol C

6.23 / 0.309 = 20 mol H

There, we have our formula PbC₈H₂₀

6 0
3 years ago
H2o, ch4, secl2, chcl3 lowest boiling point
ICE Princess25 [194]
Ch4 is the lowest boiling point
3 0
3 years ago
Use the ruler to determine the length of this object. Record your answer to the nearest tenth. The object is ____ long.
VLD [36.1K]
The object is 2.7 cm long
7 0
3 years ago
This is simple I am just confused
SOVA2 [1]

Answer:

They all have the same number of protons but different numbers of neutrons.

Explanation:

Elements will always have the same number of protons no matter the isotopes. Isotopes only change the number of neutrons. Silicon will always have 14 protons. So silicon-28 has 14 protons and 14 neutrons. Silicon-29 has 14 protons and 15 neutrons. Silicon-30 has 14 protons and 16 neutrons.

8 0
3 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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