Left Panel
The trick is to look at the protons, all other things being equal. The yellow spheres are neutrons we are led to believe. All the diagrams show the same number of electrons(2), so there is no help there.
The upper left
The upper right
The Lower left
All have 2 protons and 2 electrons. <<<< Answer Read the 3 lines above.
The center one has only 1 proton And the lower right has 3. Both of them are wrong.
Right Panel
You have answered this one correctly
Answer: 11,200 g
step by step explanation:
D= M/V
M= DV
M= (13.6 g/cm^3) (825cm^3)
M= 11,200 g
Answer:
d) The dilution equation works because the number of moles remains the same.
Explanation:
Let’s say that you have 1 mol of a solute in I L of solution. The concentration is 1 mol·L⁻¹. and <em>M</em>₁<em>V</em>₁ = 1 mol.
Now, you dilute the solution to a volume of 2 L. You still have 1 mol of solute, but in 2 L of solution. The new concentration is 0.5 mol·L⁻¹.
The volume has doubled, but the volume has halved, and <em>M</em>₂<em>V</em>₂ = 1 mol.
b) <em>Wrong</em>. The molar concentration changes on dilution.
c) <em>Wrong</em>. The volume changes on dilution.
a) <em>Wrong</em>, although technically correct, because if the moles don’t change, the mass doesn’t change either. However, the formula <em>M</em>₁<em>V</em>₁ has units mol·L⁻¹ × L = mol. Thus, in the formula, it is moles that are constant.
Answer:
I will need six (6) bags of potting soil
Explanation:
Since you plan on planting in 10 pots and need 15 cups of potting soil per pot, the total amount of potting soil you need <em>(in cups)</em> is 10 X 15 = 150 cups of potting soil.
We have that a bag contains 25 sups. to get the number of bags needed, we have to divide 150 by 25. This will give us 150 / 25 = 6 bags.
Therefore, I will need six (6) bags of potting soil
<span>Major factor which is not a part of the chemical structure that promotes the formation of solutions is the inter molecular forces between them. Also the entropy that is created in the dissolving solvent also considerably effects the solution formation</span>
