Answer:
Alpha particle
Explanation:
The helium symbol is also used to represent he alpha particles.
For example:
The americium with atomic wight 224 undergo alpha decay and produce
₉₃Np²³⁷ . The alpha particle emitted is also called helium nuclei. During this decay some gamma radiations also produce as a byproduct.
₉₅Am²²⁴ → ₉₃Np²³⁷ + ₂He⁴
Properties of alpha radiation:
Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number 4 less than and atomic number 2 less than the starting atom.
Alpha radiations can travel in a short distance.
These radiations can not penetrate into the skin or clothes.
These radiations can be harmful for the human if these are inhaled.
These radiations can be stopped by a piece of paper.
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C. An atom of helium has its valence electrons in its first energy level, it wouldn't and can't satisfy the Octet rule as it only has 2 electrons, but with 2, it has a full shell, as the first energy level can hold only 2 electrons.
Answer:
–36 KJ.
Explanation:
The equation for the reaction is given below:
2B + C —› D + E. ΔH = – 24 KJ
From the equation above,
1 mole of D required – 24 KJ of energy.
Now, we shall determine the energy change associated with 1.5 moles of D.
This can be obtained as illustrated below:
From the equation above,
1 mole of D required – 24 KJ of energy
Therefore,
1.5 moles of D will require = 1.5 × – 24 = –36 KJ.
Therefore, –36 KJ of energy is associated with 1.5 moles of D.
Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
