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tester [92]
2 years ago
5

Which of the following classification of matter includes material that can no longer be identified by their individual propertie

s?
A. Compound
B. Element
C. Mixture
D. Atom
Chemistry
2 answers:
zheka24 [161]2 years ago
8 0

Answer: Option (A) is the correct answer.

Explanation:

A substance that consists of two or more elements which are chemically combined together and cannot be distinguished by individual properties of its elements is known as a compound.

For example, water is a compound and has chemical formula H_{2}O. Water is non-flammable and non-combustible in nature whereas its elements hydrogen is extremely flammable and oxygen supports combustion.

Thus, we can conclude that in a compound classification of matter includes material that can no longer be identified by their individual properties.

xz_007 [3.2K]2 years ago
4 0
A compound consists of 2 or more elements that are combined chemically in such a way that the elements themselves can no longer be identified by their individual properties. So the Answer is A.
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Which of the following shows a Bronsted-Lowry acid reacting?
7nadin3 [17]
<h3>Answer:</h3>

         Option-C:  HCl + H₂O  →   H₃O⁺ + Cl⁻

Explanation:

       Bronsted-Lowery concept of Acid and Base defines Acid as that specie which tends to donate H⁺ (Hydrogen Ion) and bases are those species which accepts H⁺ from Acids.

In selected option, HCl is reacting as Acid as it donates H⁺ to water (lowery bronsted base).

Also, the correspong acid is converted into conjugate base (i.e. Cl⁻) and base is converted into conjugate acid (i.e. H₃O⁺)

8 0
2 years ago
Read 2 more answers
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


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Answer:

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Explanation:

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