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Tanya [424]
4 years ago
9

In which orbital does an electron in a bromine atom experience the greatest effective nuclear charge?

Chemistry
2 answers:
asambeis [7]4 years ago
4 0

First let us determine the electronic configuration of Bromine (Br). This is written as:

Br = [Ar] 3d10 4s2 4p5

 

Then we must recall that the greatest effective nuclear charge (also referred to as shielding) greatly increases as distance of the orbital to the nucleus also increases. So therefore the electron in the farthest shell will experience the greatest nuclear charge hence the answer is:

<span>4p orbital</span>

puteri [66]4 years ago
3 0

Explanation:

As the shielding effect is the effect which occurs when electrons shield each other from being attracted by the nucleus.

The effective nuclear charge is the net effective positive charge experienced by the electrons in an atom.

The electronic configuration of bromine is as follows.

      1s^{2} 2s^{2} 2^p{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{5}

Since, 1s orbital is closure to the nucleus hence it will experience the greatest nuclear charge.

You might be interested in
How many moles is ( 1 x 10^9 ) molecules O2
Effectus [21]

Answer:

The answer is

1.66 \times  {10}^{ - 15}  \:  \: moles

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L}   \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{1 \times  {10}^{9} }{6.02 \times  {10}^{23} }  \\

We have the final answer as

1.66 \times  {10}^{ - 15}  \:  \: moles

Hope this helps you

4 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
Calculate the standard free energy for the following reaction at 25°C.
sladkih [1.3K]
I dont think with this much amount of information we can solve this...unless its an reversible reaction in that case free energy =0
3 0
3 years ago
Balance the equation below and use it to answer this question:
Svet_ta [14]

Answer:

2H2 + O2 -----> 2H2O

Not sure about the second question though.

8 0
3 years ago
Maria poured 25 grams of water into a beaker. She stirred in 5 grams of salt. The salt dissolved. What is the total mass of Mari
Ahat [919]

Answer:

The answer to your question is: 30 g

Explanation:

Data

mass of water = 25 g

mass of salt = 5 g

Process

total mass = mass of water + mass of salt

                 = 25 + 5

                 = 30 g

7 0
3 years ago
Read 2 more answers
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