Answer:
31.5mL
Explanation:
The following were obtained from the question:
C1 (concentration of stock solution) = 2M
V1 (volume of stock solution) =.?
C2 (concentration of diluted solution) = 0.630M
V2 (volume of diluted solution) = 100mL
Using the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:
C1V1 = C2V2
2 x V1 = 0.630 x 100
Divide both side by 2
V1 = (0.630 x 100) /2
V1 = 31.5mL
Therefore, 31.5mL of 2M solution of FeCl2 required
1. 184.11
2. 168.812
3. 218.13
4. 272.21
1 mol is 22.4L
2.52 moles are 2.52*22.4=56.4 L
Answer:
compound 1 = 0.749gF / 1.0gl
compound 2 = 0.4489gF / 1.0gl
compound 3 = 1.047gF / 1.0gl
Explanation:
Given that;
compound mass of iodine in grams mass of fluorine in grams
1 4.75 3.56
2 7.64 3.43
3 9.41 9.86
a) Calculate the mass of fluorine per gram of iodine in each compound:
- compound 1
mass of Fluorine = (3.56gF / 4.75gF) * 1.0gl
= 0.749gF
Now RATIO = 0.749gF / 1.0gl
- compound 2
mass of Fluorine = (3.43gF / 7.64gF) * 1.0gl
= 0.4489gF
Now RATIO = 0.4489gF / 1.0gl
- compound 3
mass of Fluorine = (9.86gF / 9.41gF) * 1.0gl
= 1.047gF
Now RATIO = 1.047gF / 1.0gl
Answer:
Mineral
Explanation:
Many minerals share the same colors so it would be one of the least important parts of mineral identification
