Answer:
In order to grow a plant one should first look and compare the climate of the local environment in which the plant grows with the climate in which the person is planning to grow the plant. Thus, in the given case, Becky should do more examinations on the kind of environment and the kind of soil in which the flower is grown generally. By finding the kind of soil, one should perform some brisk assessment on the structure and composition of the soils that will permit one to find that whether or not the planting location is suitable for the growth of the flower.
Good grief, this stuff got caught in a black hole somewhere. It is terribly dense.
1 mL = 1 cc under normal conditions.
d = mass / volume
m = 20 kg
v = 5 mL
d = 20kg / 5 mL
d = 4 kg / mL
d = 4 kg / cc
A <<<<answer
Answer:
The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year
Explanation:
Given that the mass of the carbon 14 at the start = 5 gram
At the end of 5,000 years we will have;
Where
A = The amount of carbon 14 left
A₀ = The starting amount of carbon 14
e = Constant = 2.71828
= The half life
t = The time elapsed = 5000 years
λ = 0.693/ = 0.693/5730 = 0.0001209424
Therefore;
A = 5 × e^(-0.0001209424×5000) = 2.7312 grams
Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left
The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams
The average yearly rate of change of carbon-14 during the first 5000 years is therefore;
2.2688 grams/(5000 years) = 0.0004538 grams per year
The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.
Answer:
pH = 0.35
Explanation:
For a strong acid, all of the acid dissociates into H3O+, and pH = -log[H3O+], where [H3O+] = [HClO3] = 0.45 M.
Answer:
Because Iodine is a non-polar but water is polar therefore it can not have a hydrogen bond or any permanent dipole-dipole interactions
