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3 years ago

During the 1920s, installment buying, income

Chemistry

1 answer:

Neporo4naja [7]3 years ago

5 0

Answer:3

Explanation:

Not sure

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A. Identify which sets of quantum numbers are valid for an electron. Each set is ordered (n,ℓ,mℓ,ms).

Butoxors [25]

Answer:

a. Values are - (1) 3,1,-1,1/2 (2) 2,1,-1,-1/2 (3) 3,0,0,1/2 (4) 4,3,3,-1/2

(5) 3,2,2,1/2

b. Order be- I > Sn > Xe

Explanation:

Valid quantum numbers are -

'l' value should be less than 'n' value and 'ml' value should be '-l' to 'l'

'ml' value should not more than 'l' or 'n'.

Values are -

3,1,-1,1/2

2,1,-1,-1/2

3,0,0,1/2

4,3,3,-1/2

3,2,2,1/2

Given that-

I [Kr]4d¹⁰5s²5p⁵

Sn [Kr]4d¹⁰5s²5p²

Xe [Kr]4d¹⁰5s²5p⁶

Order be-

I > Sn > Xe

Xe is least because it is completely filled outer shell (5s²5p⁶

4 0

2 years ago

How much of the total mass of the solar system is contained in the Sun?

svetlana [45]

Answer:

99.8%

Explanation:

4 0

2 years ago

Read 2 more answers

Let this be an easy 5 points for your brainly account all is ask for in return is a "thank you"

Svet_ta [14]

Answer:

The number of neutrons present in one atom of isotope of Silicon of mass 28 amu is 14 neutrons

Explanation:

Symbol of Si isotope

$_{14}^{28}\textrm{Si}$

Number of Neutron = Mass number - Atomic Number

Mass number = Total number of protons and neutrons present in the nucleus of the atom.For Si = 28 amu

Atomic Number = Total number of Protons present in the nucleus.

Si = 14

Number of neutron = 24 - 14

= 14

4 0

3 years ago

Thực trạng tồn dư nitrat trên rau và an toàn thực phẩm

hram777 [196]

You are a botereleg

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4 0

2 years ago

2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2

Harrizon [31]

It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.

Explanation:

2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂

We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.

$100 g \text { of } A g N O_{3} \times \frac{2 \text { mol } A g N O_{3}}{169.87 g A g N O_{3}} \times \frac{2 \text { mol } A g C l}{1 \text { mol } A g N O_{3}} \times \frac{143.32 g A g C l}{1 \text { mol } A g C l}$

= 337.5 g AgCl

In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.

It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.

4 0

3 years ago