Answer:
4.44 M
Explanation:
The absorbance of a solution is a function of its concentration through the equation, known as Beer's law, as:
A = ε c l
where A is the absorbance, ε is is a constant of proportionality called molar absorptivity in units of L/mol·cm, c is the concentration in mol/L (M), and l is the length of the cuvette we use to measure the absorbance, usually 1 cm.
Now the above equation is a function of c in the form y= mx + b, which is exactly the linear relationship stated in the question.
Therefore
y = 0.0425 + 0.011
A = (εl) c
Therefore
slope = m = ε (1) = 0.0425
Knowing ε we can calculate the concentration of the diluted unknown:
0.200 = 0.0425 c + 0.011
c = (0.200 - 0.011)/0.425 = 0.44 M
This the concentration of the diluted solution, the original concentration of the unknown solution can be calculated from the formula V₁M₁ = V₂M₂
calling M₁ our undiluted solution, we have
M₁ = V₂M₂/V₁ = 50 mL x 0.44 M / 5.00mL = 4.44 M
(Notice there is no need to convert the volume to liters in the above equation since they appear in both sides of the equation and so they cancel each other).
Answer:
Elimination
Explanation:
Since they are removing water from the solution, it is called elimination.
Answer:
The right answer is (D)
Explanation:
If captured, some of the energy of the excited electrons is used to split carbon dioxide into carbon and oxygen.
Electron when in excited state either it jumps to the higher orbital if it has absorbed energy or it can fall to a lower orbital.
Answer:
40.68 % C
23.73 % N
8.47% H
27.12% O
Explanation:
The mass percent of an element X in a compound is calculated as the molar mass (MM) of X multiplied by the number of atoms of X in the compound, divided into the molecular weight (MW) of the compound, as follows:
mass percent of X = (MM(X) x number of atoms of X)/MW compound x 100
Thus, we first calculate the MW of acetamide (C₂H₅NO) by using the molar mass of the chemical elements C, H, N and O:
MW(C₂H₅NO) = (12 g/mol C x 2) + (1 g/mol H x 5) + 14 g/mol N + 16 g/mol O = 59 g/mol
Now, we can calculate the mass percent of each element (C, H, N, O) in C₂H₅NO:
- Mass percent of C (2 atoms of C in 1 molecule of C₂H₅NO):
% mass C = (12 g/mol x 2)/(59 g/mol) x 100 = 40.68 %
- Mass percent of N (1 atom of N in 1 molecule of C₂H₅NO):
% mass N = (14 g/mol x 1)/(59 g/mol) x 100 = 23.73 %
- Mass percent of H (5 atoms of H in 1 molecule of C₂H₅NO):
% mass H = (1 g/mol x 5)/(59 g/mol) x 100 = 8.47 %
- Mass percent of O (1 atom of O in 1 molecule of C₂H₅NO):
% mass O = (16 g/mol x 1)/(59 g/mol) x 100 = 27.12 %
The sum of the mass percents has to be equal to 100%:
40.68 % C + 23.73 % N + 8.47% H + 27.12% O = 100%