think of like a experamint and take a example of a project
Answer:
the answer is atmospheric carbon dioxide
Explanation:
The answer is 2.53e-5, I unfortunately don't know how you would really show the work other than showing the division.
Answer:
Concentration of unknown solution is 0.0416 M
Explanation:
As we know
Absorbance is equal to the product of molar absorptivity of KMnO4 m, path length and concentration
From the given set of graphical data, it is clear that the absorbance vs concentration is a straight line.
From the graph, we can obtain-
Y = 5.73 X – 0.0065
Absorbance = 0.232
0.232 = 5.73 X – 0.0065
X = 0.0416
Concentration of unknown solution is 0.0416 M
Answer:
V = 5.17L
Explanation:
Mass of gas = 8.7g
T = 23°C = (23 + 273.15)K = 296.15K
P = 1.15 atm
V = ?
R = 0.082atm.L / mol.K
From ideal gas equation
PV = nRT
P = pressure of the gas
V = volume of the gas
n = no. Of moles
R = ideal gas constant
T = temperature of the gas
no of moles = mass / molar mass
Molar mass of Chlorine = 35.5g / mol
No. Of moles = 8.7 / 35.5
No. Of moles = 0.245 moles
PV = nRT
V = nRT / P
V = (0.245 * 0.082 * 296.15) / 1.15
V = 5.9496 / 1.15
V = 5.17L
The volume of the gas is 5.17L
