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stich3 [128]
3 years ago
9

8. Which type of bond results when one or more valence

Chemistry
1 answer:
Gekata [30.6K]3 years ago
8 0

Answer:

B

Explanation:

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(20 POINTS) Easy Chemistry Question
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The correct name for the hydrocarbon would be option 2. 2 - methyl - 2 - pentene.
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A rock weighing 26.0g placed graduated cylinder displacing the volume from 13.12ml to 25.3ml what is the density of the rock in
tia_tia [17]

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3 0
3 years ago
Mass defect for oxygen-16 = 0. 13261 amu How many kilograms does the mass defect represent? 1. 66 × 10-27 kg 2. 20 × 10 -28 kg 3
Zepler [3.9K]

Mass defect for oxygen-16 = 0. 13261 amu, in the kilograms the mass defect equals to 2.20 × 10⁻²⁸ kg.

<h3>What is mass defect?</h3>

Mass defect is the difference between the mass of of an whole atom and the combined mass of its individual particles present in that atom.

We know that, 1 amu = 1.6 × 10⁻²⁷ kg

Given that, mass defect for oxygen-16 = 0.13261 amu

To calculate this defect in terms of kilograms, we have to convert into kg unit as:

0.13261 amu = 0.13261 amu × 1.6 × 10⁻²⁷ kg/amu

0.13261 amu = 2.20 × 10⁻²⁸ kg

Hence option (2) is correct.

To know more about Mass defect, visit the below link:

brainly.com/question/4334375

7 0
1 year ago
g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
3 years ago
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