Answer:
Explanation:
a. What is the mass number of the particle emitted from the nucleus during beta minus (β–) decay?
zero
The beta radiations are emitted in this reaction. The one electron is ejected and neutron is converted into proton.
⁴₆C → ¹⁴₇N + ⁰₋₁e
b. What kind of charge does the particle emitted from the nucleus during beta minus (β–) decay have?
Negative charge
Electron is emitted during beta decay and it carry negative charge.
c. What is another name for a beta minus (β–) particle?
Electron
During beta minus decay electron is emitted and neutron is converted into proton.
d. Write the balanced equation for the alpha decay that is below the “Show Equation.” Label the parent, daughter, and beta particle.
Equation is missing
a. What happens in the nucleus of an atom when an alpha particle is emitted?
When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to the starting atom.
b. b. What happens in the nucleus of an atom when a beta particle is emitted?
When nucleus emit the beta particle neutron is converted into proton and this proton stay into the nucleus while at the same time electron is emitted. Thus atomic number is increased by one.
⁴₆C → ¹⁴₇N + ⁰₋₁e
Answer:- 100
Explanations:- The given number is 124.683 and it has 6 significant figures. We are asked to round it to one significant digit number. To round it as one significant digit we need to make two second and third digits zero and all the three digits next to the decimal are dropped since we want only one significant digit. Second and third digits left to the decimal could not be dropped as they hold the place values. So, 124.683 is round to 100 that has just one significant digit.
Answer:
5 mL
Explanation:
As this is a problem regarding <em>dilutions</em>, we can solve it using the following formula:
Where subscript 1 refers to the initial concentration and volume, while 2 refers to the final C and V. Meaning that in this case:
We <u>input the data</u>:
- 5.0 M * V₁ = 0.25 M * 100 mL
And <u>solve for V₁</u>:
