Which group of nuclear emissions is listed in order of increasing charge?(1) alpha particle, beta particle, gamma radiation
(2) gamma radiation, alpha particle, beta particle
(3) positron, alpha particle, neutron
(4) neutron, positron, alpha particle
Answers
Answer: Option (4) is the correct answer.
Explanation:
An alpha particle is basically a helium nucleus and it contains 2 protons and 2 neutrons. Symbol of an alpha particle is or
.
This means that an alpha particle carries a +2 charge.
A positron is a small particle which contains a +1 charge. And, a positron is represent by the symbol .
A neutron is a sub-atomic particle present inside the nucleus of an atom. Charge on a neutron is 0.
A gamma particle is basically a photon of electromagnetic radiation with a short wavelength.
Symbol of a gamma particle is . Hence, charge on a gamma particle is also 0.
Therefore, we can conclude that group of nuclear emissions from neutron, positron, alpha particle is listed in order of increasing charge.
The group of nuclear emissions is listed in order of increasing charge is neutron, positron and alpha particle and the correct option is option 4.
Nuclear emissions refer to the particles and radiation that are emitted from the nucleus of an atom during a nuclear reaction or radioactive decay. These emissions include alpha particles, beta particles, gamma radiation, positrons, and neutrons.
- Alpha particles (α): These are helium nuclei consisting of two protons and two neutrons. They have a positive charge and relatively low penetrating power.
- Beta particles (β): These are high-energy electrons (β-) or positrons (β+) emitted during radioactive decay. Beta particles have a negative charge (β-) or positive charge (β+) and moderate penetrating power.
- Gamma radiation (γ): This is electromagnetic radiation of high energy and frequency. Gamma radiation has no charge and is highly penetrating.
- Positrons: Positrons are positively charged particles with the same mass as an electron but opposite charge. They are emitted during certain types of radioactive decay.
- Neutrons: Neutrons are electrically neutral particles found in the nucleus of an atom. They have no charge and are emitted during nuclear reactions or as a result of radioactive decay.
Thus, the ideal selection is option 4.
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To solve this we must be knowing each and every concept related to mass. Therefore, mass of KSO
would you need to prepare 1500g of 5.0% K
SO
(m/m) solution is 75 g.
What is mass?
Mass defines the quantity of a substance. It is measured in gram or kilogram. Average mass is the mass of atoms of an element that are isotopes. It can be calculated by multiplying mass of a isotope to natural abundance of that isotope.
Mass of solution = 1500 g
5.0 % = 5.0 / 100 = 0.05
mass percentage = mass of solute / mass of solution
substituting all the given values in the above equation, we get
0.05 = mass of solute / 1500
mass of solute = 0.05 x 1500
mass of solute = 75 g of solute
Therefore, mass of KSO
would you need to prepare 1500g of 5.0% K
SO
(m/m) solution is 75 g.
To learn more about mass, here:
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Mass of solution = 1500 g
5.0 % = 5.0 / 100 = 0.05
w = mass of solute / mass of solution
0.05 = ? / 1500
w = 0.05 x 1500
w = 75 g of solute
hope this helps!
Answers
Rate of acceleration of the cup due to force of wind = 5 m/s^2
We already know that
Force = Mass * Acceleration
So Force exerted by the wind on the cup = 0.025 * 5 kg m/s^2
= 0.125 Newton
So the force exerted by the wind on the paper cup is 0.125 newton.I hope this is the answer thatyou wanted and the process of doing such problems is clear to you.
Answers
Answers
A car traveling at 20 m/s starts to decelerate steadily it comes to a complete stop in 10s. The equation for the acceleration is velocity divided by time. So divide 20 m/s by 10 s and you will get an acceleration of 2 m/s2.
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1) Assume 100 g sample
63.15% C * 100 g = 63.15g
5.30% H * 100 g = 5.30g
31.55% O * 100g = 31.55g
2) Convert mass to moles using their atomic weights
63.15 g * 1 mol C / 12.0107 g C = 5.2870 mol C
5.30 g * 1mol H / 1.0079 g H = 5.2585 mol H
31.55 g * 1mol O / 15.9994 O = 1.9719 mol O
3) Divide each quantity by the smallest number of moles
5.2870 mol C / 1.9719 mol = 2.6812 C = 2 C
5.2584 mol H / 1.9719 mol = 2.6667 H = 2 H
1.9719 mol O / 1.9719 mol = 1.000 O = 1 O
Empirical Formula is C₂H₂O
If the problem is silent, the molecular weight is equivalent to the empirical weight.
To get the molecular formula, divide the molecular weight by the empirical weight to get the multiple.
Molecular Weight is not mentioned thus it is equivalent to Empirical Weight which is:
Atom Number in Molecule Atomic Weight Total Mass
C 2 12.0107 24.0214
H 2 1.0079 2.0158
O 1 15.9994 15.9994
Total weight is 42.0366
Molecular weight / Empirical Weight = Multiple to be multiplied to the Empirical Formula
42.0366 / 42.0366 = 1
Molecular Formula isC₂H₂O
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Benzene
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Benzopyrene
Ethenone
Acetic acid
Methane
Butane
Heptane
Pentane