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valkas [14]
3 years ago
15

3 What is the number located in the top right hand corner of the atomic symbol of an ion with 16 protons, 17 neutrons and 14 ele

ctrons? * (1 Point) +1 +2 -2 0 -1 Submit​
Chemistry
1 answer:
siniylev [52]3 years ago
3 0

The number located in the top right corner of the ion is +2

An ion is an atom or group of atoms which possess an electric charge.

The charge on an ion can be obtained by using the following formula:

<h3>Charge = Proton – Electron </h3>

With the above formula, we can obtain the charge of the ion given in the question above as illustrated below:

Proton = 16

Neutron = 17

Electron = 14

<h3>Charge =? </h3>

Charge = Proton – Electron

Charge = 16 – 14

<h3>Charge = +2</h3>

Therefore, the charge on the ion is +2

Learn more: brainly.com/question/3428265

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Branches of chemistry with definition and examples in real life
nataly862011 [7]

Explanation:The five major branches of chemistry are organic, inorganic, analytical, physical, and biochemistry.

...

Sub-branches of physical chemistry include:

Photochemistry — the study of the chemical changes caused by light.

Surface chemistry — the study of chemical reactions at surfaces of substances

8 0
3 years ago
2) 2KClO3 --&gt; 2KCl + 3O2
aleksandrvk [35]

2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2

a)

2 \text{ mols of KClO}_3 \equiv 3  \text{ mols of O}_2

19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5  \text{ mols of O}_2

\boxed{19 \text{ mols of KClO}_3 \equiv 28,5  \text{ mols of O}_2}

b)

2 \text{ mols of KClO}_3 \equiv 2  \text{ mols of KCl}

62 \text{ mol of KClO}_3 \equiv 62  \text{ mol of KCl}

Using the atomic mass given in the periodic table:

62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

62\cdot122,5 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

7595 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

\boxed{7,595 \text{ kg of KClO}_3 \equiv 62  \text{ mol of KCl}}

c)

2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3

3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}

Using the atomic mass given in the periodic table:

3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5)  \text{ g of KCl}

96\text{ g of O}_2 \equiv 149\text{ g of KCl}

\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}

\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}

This result is an aproximation.

8 0
3 years ago
In the reaction 2Na + Cl2 → 2NaCl, the products are
Lisa [10]
During a reaction, there are reactants and products. Reactants are the elements that are being added. Products are the result of the reactants.

2Na + Cl2 are the reactants.

2NaCl is the Product.

I hope this helps!
8 0
3 years ago
Instrument for measuring the basic units​
o-na [289]

Answer: A Ruler

Explanation:A Ruler is a common instrument used for measuring the length of small objects. It usually has four units of measurement to choose from: millimeter, centimeter, inch, and foot.

4 0
3 years ago
Which of the following represent impossible combinations of n and I? (a) 1p, (b) 4s, (c) 5f, (d) 2d 6.61
Eduardwww [97]

Answer:

  • <u>Options (a) and (d) : (a) 1p and (d) 2d represent impossible combinations of n and l.</u>

Explanation:

n refers to the principal quantum number, and l refers to the angular momentum or Azimuthal quantum number.

Principal quantum number (n) is used to indicate the main energy level of the electron. It may take whole numbers: 1, 2, 3, 4, 5, 6, 7, ...

Angular momentum or Azimuthal quantum number refers to the kind (shape) of the orbital. It can take numbers from 0 to n - 1.

So, if n = 1, l can only be 0; if n = 2, l can be either 0 or 1; if n = 2, l can be either 0, 1 or 2.

On the other hand, the shape of the orbitals is also representd by a letter. then there is a unique relation between the letter that represents the orbital and the angular quantum number which is:

letter       l number

s               0

p               1

d               2

f                3

The previous information is summarized in the next table:

n    possible l numbers          

1          0                                          

2         0, 1                                        

3         0, 1, 2                                    

4         0, 1, 2, 3                              

5        0, 1, 2, 3, 4

6        0, 1, 2, 3, 4, 5

7        0, 1, 2, 3, 4, 5, 6

As per the choices given in the question you have:

<u>(a) 1 p</u> is not possible because when n = 1 the only l number is 0 and it is an s orbital, but p ⇒ l = 1. Thus, this is a correct choice.

<u>(b) 4s</u> is possible since n = 4 permits l to be 0, 1, 2, and 3.

<u>(c) 5f </u>is possible since n = 5 permits l to be 0, 1, 2, 3, and 4.

<u>(d) 2d</u> is impossible since n = 2 permits l to be 0, and 1, but d ⇒ l = 2. Thus, this is other right choice.

4 0
3 years ago
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