I think the answer is c not completely sure
The question is incomplete.
You need two additional data:
1) the original volume
2) what solution you added to change the volume.
This is a molarity problem, so remember molarity definition and formula:
M = n / V in liters: number of moles per liter of solution
To give you the key to answer this kind of questions, supppose the original volumen was 1 ml and that you added only water (solvent).
The original solution was:
V= 1 ml
M = 0.2 M
Using the formula for molarity, M = n / V
n = M×V = 0.2 M × (1 / 10000)l = 0.0002 moles
For the final solution:
n = 0.0002 moles
M = 0.04
From M = n / V ⇒ V = n / M = 0.002 moles / 0.04 M = 0.05 l
Change to ml ⇒ 0.05 l × 1000 ml / l = 50 ml. This would be the answer for the hypothetical problem that I assumed for you.
I hope this gives you all the cues you need to answer similar problems about molarity.
Answer:
The ring of fire
Explanation:
The ring of fire is where they meet or at the fault lines
Answer:
23.34 %.
Explanation:
- The percentage of water must be calculated as a mass percent.
- We need to find the mass of water, and the total mass in one mole of the compound. For that we need to use the atomic masses of each element and take in consideration the number of atoms of each element in the formula unit.
- <em>Atomic masses of the elements:</em>
Cd: 112.411 g/mol, N: 14.0067 g/mol, O: 15.999 g/mol, and H: 1.008 g/mol.
- <em>Mass of the formula unit:</em>
Cd(NO₃)₂•4H₂O
mass of the formula unit = (At. mass of Cd) + 2(At. mass of N) + 10(At. mass of O) + 8(At. mass of H) = (112.411 g/mol) + 2(14.0067 g/mol) + 10(15.999 g/mol) + 8(1.008 g/mol) = 308.5 g/mol.
- <em> Mass of water in the formula unit:</em>
<em>mass of water</em> = (4 × 2 × 1.008 g/mol) + (4 × 15.999 g/mol) = 72.0 g/mol.
- <em>So, the percent of water in the compound = [mass of water / mass of the formula unit] × 100 = [(72.0 g/mol)/(308.5 g/mol)] × 100 = 23.34 %</em>
if there is no carbon dioxide your test tube will be blue
if there is a medium amount of carbon dioxide your test tube is green
if there are high amounts of CO2 it will be
yellow
