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olganol [36]
3 years ago
15

He partial lewis structure that follows is for a hydrocarbon molecule. in the full lewis structure, each carbon atom satisfies t

he octet rule, and there are no unshared electron pairs in the molecule. the carbon-carbon bonds are labeled 1, 2, and 3. part a how many hydrogen atoms are in the molecule?
Chemistry
2 answers:
Bezzdna [24]3 years ago
6 0

Answer:

Four (4) hydrogen atoms

Explanation:

C=C-C(triple bond)C

only four hydrogen atoms can bond with the vacant space

ziro4ka [17]3 years ago
4 0

Answer:

4 hydrogen atoms are attached with the carbon atom in the molecule.

Explanation:

Hydrocarbon molecule is a type of molecule which is composed of hydrogen and carbon atoms. Carbon atom belongs to group 4 in the periodic table which means it has four electrons in its outermost shell or orbit, so for attaining octet rule, carbon atom makes 4 covalent bonds with 4 hydrogen atoms and 4 electrons of hydrogen atom is used by the carbon atom and becomes a stable molecule.

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If an object is accelerating the forces acting on the object are ?​
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7 0
3 years ago
25cc of 5 % NaOH solution neutralized 30cc of h2sO4 solution. Whatis normality of H2SO4?
MAVERICK [17]

The normality of the H₂SO₄ that reacted with 25cc of 5 % NaOH solution is 1.1 N.

<h3>What is the molarity of 5% NaOH?</h3>

The molarity of 5% NaOH is 1.32 M

25 cc of NaOH neutralized 30cc of H₂SO₄ solution.

Equation of reaction is given below:

  • 2 NaOH + H₂SO₄ ---> Na₂SO₄ + 2 H₂O

Molarity of H₂SO₄ = 1.32 x 1 x 25/(30 x 2) = 0.55 M

  • Normality = Molarity × moles of H⁺ ions per mole of acid

moles of H⁺ ions per mole of H₂SO₄ = 2

Normality of H₂SO₄ = 0.55 x 2 = 1.1 N

In conclusion, the normality of an acid is determined from the molarity and the moles of H⁺ ions per mole of acid.

Learn more about normality at: brainly.com/question/22817773

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7 0
2 years ago
A 0.300 m solution of hcl is prepared by adding some 1.50 m hcl to a 500 ml volumetric flask and diluting to the mark with deion
liberstina [14]
In dilution we add distilled water to decrease the concentration of required sample from high concentration to lower concentration
The law used for dilution:
M₁V₁]Before dilution = M₂V₂] After dilution
M₁ = 1.5 M
V₁ = ?
M₂ = 0.3 M
V₂ = 500 ml
1.5 * V₁ = 0.3 * 500 ml
so V₁ = 100 ml and it completed to 500 ml using 400 ml deionized water
8 0
3 years ago
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