Answer:
Part A. 1.355 mol/L
Part B. 0.100 mol
Part C. 74.0 mL
Explanation:
Part A.
The molar mass of luminol is 177.16 g/mol, so the number of moles at 18.0 g is:
n = mass/molar mass
n = 18.0/177.16
n = 0.1016 mol
The molarity is the number of moles divided by the volume (0.075 L)
C = 0.1016/0.075
C = 1.355 mol/L
Part B.
The number of moles is the molarity multiplied by the volume, so:
n = 5.00x10⁻² mol/L * 2.00 L
n = 0.100 mol
Part C.
To prepare a solution by dilution, we can use the equation
C1V1 = C2V2
Where C1 is the concentration of the initial (stock) solution, V1 is its volume necessary, C2 is the concentration of the diluted solution, and V2 is its volume.
Thus, C1 = 1.355 M, C2 = 0.05 M, V2 = 2.00 L
1.355V1 = 0.05*2
V1 = 0.074 L
V1 = 74.0 mL
Answer:
0.198 g CO2 plz mark me brainliest
Explanation:
Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons
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Answer:
![\frac{[magenta\ phenolphthalein]}{[colorless\ phenolphthalein]}=31.62](https://tex.z-dn.net/?f=%5Cfrac%7B%5Bmagenta%5C%20phenolphthalein%5D%7D%7B%5Bcolorless%5C%20phenolphthalein%5D%7D%3D31.62)
Explanation:
Considering the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution as:
![pH=pKa+\log\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2B%5Clog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
Where Ka is the dissociation constant of the acid.
pKa of phenolphthalein = 9.40
pH = 10.9
So,
![10.9=9.40+\log\frac{[magenta\ phenolphthalein]}{[colorless\ phenolphthalein]}](https://tex.z-dn.net/?f=10.9%3D9.40%2B%5Clog%5Cfrac%7B%5Bmagenta%5C%20phenolphthalein%5D%7D%7B%5Bcolorless%5C%20phenolphthalein%5D%7D)
