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KiRa [710]
3 years ago
13

What particles contribute to the mass of an atom?

Chemistry
2 answers:
ira [324]3 years ago
6 0
Protons and Neutrons
viva [34]3 years ago
6 0

Answer:

Protons and Neutrons

Explanation:

Since the nucleus of an atom contribute to the mass of atom and protons and neutrons are present inside so proton and neutrons contribute to the mass of an atom. Electron is not so heavy particle so it won't contribute to the mass of an atom.

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(20 Points)
kogti [31]
A. It is Basic

for example,

NaOH → Na⁺(aq) + OH⁻(aq)
Ba(OH)₂ → Ba²⁺(aq) + 2OH⁻(aq)
Ca(OH)₂ → Ca²⁺(aq) + 2OH⁻(aq)

pH>7
6 0
3 years ago
Read 2 more answers
4.
Evgen [1.6K]

Answer:

λ = 2.8 m

Explanation:

Given data:

Frequency of radio wave = 106.7 ×10⁶ Hz

Wavelength of radio wave = ?

Solution:

Formula:

Speed of wave = frequency  × wavelength

speed of wave = 3×10⁸ m/s

by putting values,

3×10⁸ m/s = 106.7 ×10⁶ Hz × λ

Hz = s⁻¹

λ = 3×10⁸ m/s / 106.7 ×10⁶ Hz

λ =  3×10⁸ m/s / 106.7 ×10⁶ s⁻¹

λ =  0.028×10² m

λ = 2.8 m

7 0
2 years ago
What does the atomic numbrr represent on the periodic table?​
lesya692 [45]

Answer:

Explanation:

The atomic number tell us the number of protons and neutrons in the nucleus of an atom. in other words ,each element has a unique number that identifies how many protons are in one atom of that element example, all hydrogen atoms, and only hydrogen atoms, contain one proton and have an atomic number of 1.

4 0
3 years ago
On the periodic table, what is used to determine the order of elements? (Use your own words? Help please!!!
Angelina_Jolie [31]

Elements are listed in order of increasing atomic number from the left to the right.

4 0
3 years ago
Read 2 more answers
5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wav
torisob [31]

Answer: The range of wavelengths of light that can be used to cause given phenomenon is 8.953 \times 10^{21} m.

Explanation:

Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J

Formula used is as follows.

E = \frac{hc}{\lambda}

where,

E = energy

h = Planck's constant = 6.625 \times 10^{-25} Js

c = speed of light = 3 \times 10^{8} m/s

Substitute the values into above formula as follows.

E = \frac{hc}{\lambda}\\222000 J = \frac{6.625  \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m

Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is 8.953 \times 10^{21} m.

7 0
3 years ago
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