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liq [111]
3 years ago
15

What is true about an atom in an excited state? Select all that apply:

Chemistry
1 answer:
AnnyKZ [126]3 years ago
4 0

Explanation:

What is true about an atom in an excited state? Select all that apply:

A. It is less likely to enter chemical reactions.

B. It is more likely to enter chemical reactions.

C. It has less energy than a ground state atom.

D. It has more energy than a ground state atom.

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A mover packs books, CDs, and DVDs into a moving box. If the box contains 6.5 kg of books, 1.5 kg of CDs, and 2.0 kg of DVDs, wh
mariarad [96]
Given: 
<span>M1 = 6.5 kg of books
</span><span>M2 = 1.5 kg of CDs
</span><span>M3 = 2.0 kg of DVDs

Required: percent by mass of each object

Solution:
First, we calculate the total mass.

M = 6.5 kg + 1.5 kg + 2.0 kg =  10 kg

Percent by mass is calculated by getting the ration of the mass of an object and the total mass multiplied by 100 to get the percent.

%M1 = 6.5 / 10 x 100 = 65%
%M2 = 1.5/10 x 100 = 15%
%M3 = 2.0/10 x 100 = 20%</span>
4 0
3 years ago
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8 0
3 years ago
The electron configuration of an element is 1s22s22p63s1.
k0ka [10]

The element is Sodium with an atomic number of 11 and electrovalent bonding takes place when it comes near an atom having seven valence electrons.

<h3>What is Electrovalent bonding?</h3>

This is also referred to as ionic bonding and involves the transfer of atoms of an element to another.

In order for both of them to achieve a stable octet configuration, sodium donates one atom to the element seven valence electrons.

Read more about Electrovalent bonding here brainly.com/question/1979431

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6 0
2 years 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⁻

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