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ahrayia [7]
3 years ago
8

Cindy dissected a flowering plant and looked at the stem, roots, leaves, and flower under a microscope. After her observations,

Cindy should conclude that all flowering plant structures __________.
are not living


have the same function


are green


have cells
Chemistry
1 answer:
yarga [219]3 years ago
4 0
D. Have cells. :) :)

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Gallium chlorate decomposes
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Explanation:

in the decomposition of gallium chlorate, 75.0grams is heated to produce oxygen and gallium chloride.

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Why do graphite and diamond appear to be different substances if they are composed of the same atom?
Alekssandra [29.7K]

Answer:

Diamond and graphite are allotropes of carbon.

Explanation:

Diamond has a close packed crystal structure which is responsible for its extreme hardness. In it, each carbon atom is sp³ hybridised and bonded to four other carbon atoms in a tetrahedral arrangement. Diamond has a hardness of 10 on mohs scale with a cubic crystal form.

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Graphite on the other hand is made up of layers of hexagonal structure that are weakly bonded by van-der-waals forces. This layered arrangement explains its softness/slippery feeling and hence it is used as a lubricant. In each layer, each carbon atom is sp² hybridised and bonded to three other carbon atoms in a trigonal planar arrangement.

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Pine trees slightly prefer slightly acidic soil. What happens if the pH changes to above 7?​
egoroff_w [7]

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The trees' growth might be affected

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Read 2 more answers
Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+
DIA [1.3K]

These are four questons and four answers:

Answers:

  • 1)  7⁺
  • 2) 1⁻
  • 3) 3⁺
  • 4) 5⁺

Explanation:

<u><em>Question 1) </em></u><u><em>Cl₂O₇:</em></u>

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

  • 2×X + 7 (-2) = 0
  • 2X - 14 = 0
  • 2X = +14
  • X = +14 /2 = + 7

<em>Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.</em>

<u><em>Question 2) </em></u><u><em>AlCl₄⁻</em></u>

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

  • 1 (+3) + 4X = - 1
  • +3 + 4X = - 1
  • 4X = - 1 - 3
  • 4X = - 4
  • X = - 1

<em>Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.</em>

<em><u>Question 3)</u></em><em><u> Ba(ClO₂)₂</u></em>

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

  • +2 + 2X + 4 (-2) = 0
  • 2X +2 - 8 = 0
  • 2X - 6 = 0
  • 2X = +6
  • X = + 3

<em>Conclusion: the oxidation number of Cl in Ba(ClO₂)₂  is 3⁺.</em>

<u><em>Question 4)</em></u><u><em> CIF₄⁺</em></u>

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

  • X + 4(-1) = +1
  • X - 4 = +1
  • X = +1 + 4
  • X = + 5

<em>Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.</em>

6 0
3 years ago
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