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Alex Ar [27]
3 years ago
12

Which subatomic particles have approximately the same mass?

Chemistry
2 answers:
igor_vitrenko [27]3 years ago
8 0

Answer:

The 3rd one is correct

Explanation:

Protons and neutrons have approximately the same mass, about 1.67 × 10−24 grams, which scientists define as one atomic mass unit (amu) or one Dalton. Each electron has a negative charge (−1) equal to the positive charge of a proton (+1). Neutrons are uncharged particles found within the nucleus.

zvonat [6]3 years ago
5 0

Explanation:

protons and neutrons have approximately the same size, the electron is 1847th, (I think), the size of a proton

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What is the number of atoms in 3.75 mol fe
Kruka [31]
1mol=6.022x10^23 atoms
3.75mol= 6.022x3.75x10^23 atoms
             =2.2583x10^24 atoms
8 0
3 years ago
A sample of metal has a mass of 16.5916.59 g, and a volume of 5.815.81 mL. What is the density of this metal
Oxana [17]

Answer:

=> 2.8554 g/mL

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m)     = 16.59 g

Volume (v) =  5.81 mL

From our question, we are to determine the density (rho) of the rock.

The formula:

p = \frac{m}{v}

Substitute the values into the formula:

​p = \frac{16.59 g}{5.81 mL} \\   = 2.8554 g/mL

= 2.8554 g/mL

Therefore, the density (rho) of the rock is 2.8554 g/mL.

5 0
2 years ago
Which of the following is true of solids?
Over [174]
<span>Answer: B. Ionic solids have higher melting points than molecular solids.
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5 0
3 years ago
(The radioisotope 224Ra decays by alpha emission via two paths to the ground state of its daughter 94% probability of alpha deca
pav-90 [236]

Answer:

a) ²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q

b) the Q-value of this reaction is 5.789 MeV

c) the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev

       

Explanation:

a)

The decay equation for the alpha decay is expressed as;

²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q

b)

Calculate the Q-value (in MeV) of this reaction.

Q = Mparent - Mdaughter -Mg

Q = MRa - MRn -Mg

= 224.020202 - 220.011384 - 4.00260305

= 0.00621495 amu

= 5.789 MeV

therefore the Q-value of this reaction is 5.789 MeV

c)

Energy of alpha particle is expressed as;

E∝ = MQ / ( m + M)

now this is the maximum energy available for the daughter, ²²⁰Rn going to the ground state;

The energy of the alpha particle gives;

E∝  = 220(5.789) / ( 4 + 220) = 5.69 MeV

as given in the question,The other less frequent alpha occurring 5.5% of the time leaves the daughter nucleus in an excited state of 0.241 MeV above the ground state.

Therefore the energy of this alpha is

E∝ = 5.69 - 0.241 = 5.449 Mev

Therefore the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev

d)

Sketch of the nuclear decay scheme have been uploaded along side this answer.

7 0
3 years ago
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Aleksandr-060686 [28]

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3 0
3 years ago
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