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Alisiya [41]
3 years ago
7

A certain atom has 22 protons and 19 electrons. This atom loses an electron. The net charge on the atom is now . If this same at

om with 22 protons and 19 electrons were to gain 3 electrons, the net charge on the atom would be .
Chemistry
1 answer:
ASHA 777 [7]3 years ago
6 0

1) p⁺ = 22; number of protons.

e⁻ = 19 - 1 = 18; number of electrons.

Net charge is +4, because atom has 4 protons more than electrons.

Proton is a subatomic particle with a positive electric charge of +1e elementary charge.

2) p⁺ = 22; number of protons.

e⁻ = 19 + 3 = 22; number of electrons.

Net charge is 0 (neutral charge), because atom has same number of protons and electrons.

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Number of electrons for all of these even the ones with a number by them ? Will give brain-list!!!!
DaniilM [7]

Explanation:

s=2

p=6

d=10

f=14

I hope this was helpful

5 0
3 years ago
Consider the reaction:
Fofino [41]

The mass in grams of NH₃ produced from the reaction is 3.4 g

<h3>Balanced equation</h3>

We'll begin by writing the balanced equation for the reaction. This illustrated below:

N₂ + 3H₂ -> 2NH₃

From the balanced equation above,

1 dm³ of N₂ reacted to produced 2 dm³ NH₃

<h3>How to determine the volume of NH₃ produced</h3>

From the balanced equation above,

1 dm³ of N₂ reacted to produced 2 dm³ NH₃

Therefore,

2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃

<h3>How to determine the mass of NH₃ produced</h3>

We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:

22.4 dm³ = 1 mole NH₃

Therefore,

4.48 dm³ = 4.48 / 22.4

4.48 dm³ = 0.2 mole of NH₃

Finally, we shall determine the mass of NH₃ as follow:

  • Molar mass of NH₃ = 17 g/mol
  • Mole of NH₃ = 0.2 mole
  • Mass of NH₃ =?

Mass = mole × molar mass

Mass of NH₃ = 0.2 × 17

Mass of NH₃ = 3.4 g

Learn more about stoichiometry:

brainly.com/question/13196642

#SPJ1

4 0
1 year ago
1. The red light associated with the aurora borealis is emitted by excited oxygen atoms. This
stiv31 [10]

Answer:

Frequency = 0.005× 10¹⁷ s⁻¹

E =  0.03 × 10⁻²⁰KJ

Explanation:

Given data:

Wavelength of light = 630.0 nm (630 × 10⁻⁹ m)

Frequency of light = ?

Energy in joule = ?

Solution:

speed of light = wavelength × frequency

Frequency = speed of light /  wavelength

Frequency = 3 × 10⁸ m/s / 630 × 10⁻⁹ m

Frequency = 0.005× 10¹⁷ s⁻¹

Energy:

E = h. f

E = 6.63× 10⁻³⁴ Kg. m²/s × 0.005× 10¹⁷ s⁻¹

E = 0.03 × 10⁻¹⁷ j

E = 0.03 × 10⁻¹⁷ j×1 kj /1000 j

E =  0.03 × 10⁻²⁰KJ

4 0
3 years ago
How are isotopes defined? forms of different elements that have the same number of neutrons in each atom forms of different elem
Gre4nikov [31]

Answer : The correct option is, forms of the same element that differ in the number of neutrons in each atom

Explanation :

Isotopes : It is defined as the forms of same element that have the same number of protons and electrons but differ in the number of neutrons.

For example : Carbon-12, carbon-13 and carbon-14 is an isotopes due to the same number of protons and electrons and different number of neutrons.

Carbon-12

Atomic number = 6

Atomic mass = 12

Atomic number = Number of protons = Number of electrons = 6

Number of neutrons = Atomic mass - Atomic number = 12 - 6 = 6

Carbon-13

Atomic number = 6

Atomic mass = 13

Atomic number = Number of protons = Number of electrons = 6

Number of neutrons = Atomic mass - Atomic number = 13 - 6 = 7

Carbon-14

Atomic number = 6

Atomic mass = 14

Atomic number = Number of protons = Number of electrons = 6

Number of neutrons = Atomic mass - Atomic number = 14 - 6 = 8

3 0
4 years ago
Read 2 more answers
Given that a 11.00 g milk bar chocolate bar contains 7.500 g of sugar, calculate the percentage of sugar present in 11.00 g of m
alukav5142 [94]

Answer:

68.18%

Explanation:

The question asking for the percentage of sugar present in milk bar chocolate. To find the percentage of a certain molecule, you have to divide the mass of the molecule with the total mass of the product. The calculation will be:

sugar percentage = sugar weight / milk bar chocolate weight * 100%

sugar percentage = 7.5g / 11g * 100%= 68.18%

7 0
3 years ago
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