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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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The energy produced when nucleons fuse together is called the: Select the correct answer below:
saveliy_v [14]

Answer:

Option C (nuclear binding energy) is the appropriate choice.

Explanation:

  • At either the nuclear scale, the nuclear binding energy seems to be the energy needed to remove and replace a structure of the atom itself into the characterize elements (to counteract the intense nuclear arsenal).
  • Nuclear warheads (bargaining power) bind everything together neutrons as well as protons within an elementary particle.

Some other options in question aren't relevant to the particular instance. So that the option preceding will also be the right one.

7 0
3 years ago
What is the power produced by a light bulb with a voltage of 120 volts and a current of 0.045 amps? PLEASE HELP I HAVE 5 MINUTES
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The answer is 5.4! I found it on a an online calculator! 
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A white salt containing an unknown metal has the formula MCl and gives a lilac flame during a flame test. the salt could be
Tatiana [17]

Answer:

C

Explanation:

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2 years ago
Is O2 covalent molecular or polar covalent
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The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ rad
Yuliya22 [10]

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = x\times \frac{56.4}{100}=(0.564x)pm

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}

2\times x+2\times 0.564x=566

2x+1.128x=566

3.128x=566

x=180.9\approx 181pm

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

4 0
3 years ago
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