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Alexandra [31]
2 years ago
12

Start four fundamental laws of chemical Combinations and the scientist that stated each below (1) Law of Conseriation of mass (2

) Law of Definite proporsionis (3)Law of multiple proportions (4)Law of reciprocal propertion​
Chemistry
1 answer:
kirza4 [7]2 years ago
8 0

Explanation:

Law of conservation of matter/mass

the law of conservation of mass states that matter cannot be created nor destroyed but can be changed from one form to another.

by Antoine Lavoisier

Law of multiple proportions

the law of multiple proportions states that if two elements, A and B combine to form more than one chemical compound, the various masses of element A, which combine separately with a fixed mass of the other element B are in simple multiple ratios.

by chemist John Dalton

Law of reciprocal proportion

the law of reciprocal proportion states that the masses of several elements A, B, C

which combine separately with a fixed mass of another element D are the same as or simple multiples of, the masses in which A, B, and C themselves combine.

by Jeremias Richter

Law of definite proportions

the law of definite proportions states that all pure samples of a particular chemical combined contain similar elements combined in the same proportion by mass.

by Joseph Proust

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What volume, in milliliters, of a 0.997 M KOH solution is needed to neutralize 30.0 mL of 0.0400 M HCl?
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Answer:

1.2 mL

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

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From the balanced equation,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of base, KOH (n₆) = 1

Finally, we shall determine the volume of the base, KOH needed to neutralize the acid, HCl as follow:

Molarity of base, KOH (M₆) = 0.997 M

Volume of acid, HCl (Vₐ) = 30 mL

Molarity of acid, HCl (Mₐ) = 0.0400 M

Volume of base, KOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ/n₆

0.04 × 30 / 0.997 × V₆ = 1/1

1.2 / 0.997 × V₆ = 1

Cross multiply

0.997 × V₆ = 1.2

Divide both side by 0.997

V₆ = 1.2 / 0.997

V₆ = 1.2 mL

Thus, the volume of the base, KOH needed to neutralize the acid is 1.2 mL.

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