Answer: B) metals, non-metals, metalloids
An example of a metal is iron. A non-metal example is oxygen, which is a gas at STP (standard temperature and pressure).
A metalloid is a bit of a mix between a metal and non-metal element. It's sorta like an element that has both properties of metals and non-metals, or it's in a murky gray area. An example of a metalloid would be silicon.
Answer:
Here's how I would explain it.
Explanation:
Think of it this way.
When you mix solutions of silver nitrate and sodium chloride, you get an immediate precipitate of silver chloride. The equation is
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
Now, take some AgCl and stir it vigorously with water.
You won't see much happening, because the AgCl is has such a low solubility. Not much of it will go into solution. And yet, a small amount of it does dissolve until the solution is saturated.
The concentration of AgCl in the saturated solution is
about 0.000 01 mol·L⁻¹.
I hope you will agree that this is a dilute solution even though it is saturated with AgCl.
The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g
The equation is Mr=Mi(1/2)^n
where n is the number of half-lives
Mr is the mass remaining after n half lives
Mi is the initial mass of the sample
To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)
n=1000/24100=0.0414
So Mr=5x(1/2)^1=2.5 g
The mass remaining is 2.5 g
- The half life is the time in which the concentration of a substance decreases to half of the initial value.
Learn more about half life at:
brainly.com/question/24710827
#SPJ4
Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O
Basically the sugar breaks down until it’s eventually evaporated and spreads to different parts of the water solution and while it’s spreading the chemicals and the flavors in the sugar are going into the molecules and atoms in the water and mixing to make the water solution taste sweet just like sugar. So i would say it would be a physical change not a chemical change.