Answer:
Potential energy = 441 N
Explanation:
Given:
Mass M = 15 kg
Height = 3 m
Find:
Potential energy
Computation:
Potential energy = mgh
Potential energy = (15)(9.8)(3)
Potential energy = 441 N
Jellyfish reproduction<span> involves several different stages. In the adult, or medusa, stage of a </span>jellyfish<span>, they can </span>reproduce<span>sexually by releasing sperm and eggs into the water, forming a planula. ... During this stage, which can last for several months or years, asexual </span>reproduction<span> occurs.</span>
Answer:
O₂; KCl; 33.3
Explanation:
We are given the moles of two reactants, so this is a limiting reactant problem.
We know that we will need moles, so, lets assemble all the data in one place.
2KCl + 3O₂ ⟶ 2KClO₃
n/mol: 100.0 100.0
1. Identify the limiting reactant
(a) Calculate the moles of KClO₃ that can be formed from each reactant
(i)From KCl
(ii) From O₂
O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.
KClO₃ is the excess reactant.
2. Moles of KCl left over
(a) Moles of KCl used
(b) Moles of KCl left over
n = 100.0 mol - 66.67 mol = 33.3 mol
Answer:
Helium nuclei is emitted during alpha decay.
Explanation:
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.
Properties of alpha radiation:
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.
Example:
₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy
