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Answer:
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Nitrogen forms many thousands of organic compounds. Most of the known varieties may be regarded as derived from ammonia, hydrogen cyanide, cyanogen, and nitrous or nitric acid. The amines, amino acids, and amides, for example, are derived from or closely related to ammonia.
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Explanation:
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A carbohydrate has only Carbon, Hydrogen and Oxygen, with the hydrogen and oxygen in the ratio 2:1.
Examples are glucose C6H12O6 and table sugar C12H22O11
A protein is made of amino acids. Amino acids have an amino group which is −NH2.
So without nitrogen atoms the compound cannot be a protein.
Nucleic acids are organic compounds that contain carbon, hydrogen, nitrogen but they also contains phosphorus and oxygen. They are made of smaller units called nucleotides.
So;
Simple Amines are the class of organic compounds that only contains Carbon, Hydrogen and Nitrogen.
Answer:
Explanation:
Given parameters :
Volume of solution = 100mL
Absorbance of solution = 0.30
Unknown:
Concentration of CuSO₄ in the solution = ?
Solution:
There is relationship between the absorbance and concentration of a solution. They are directly proportional to one another.
A graph of absorbance against concentration gives a value of 0.15M at an absorbance of 0.30.
The concentration is 0.15M
Also, we can use: Beer-Lambert's law;
A = ε mC l
where εm is the molar extinction coefficient
C is the concentration
l is the path length
Since the εm is not given and assuming path length is 1;
Then we solve for the concentration.
Answer:
M(Fe₂O₃) = 159.70 g/mol
M(CO) = 28.01 g/mol
M(Fe) = 55.85 g/mol
M(CO₂) = 44.01 g/mol
Explanation:
We can calculate the molar mass of a compound by summing the molar masses of the elements that form it.
Fe₂O₃
M(Fe₂O₃) = 2 × M(Fe) + 3 × M(O) = 2 × 55.85 g/mol + 3 × 16.00 g/mol = 159.70 g/mol
CO
M(CO) = 1 × M(C) + 1 × M(O) = 1 × 12.01 g/mol + 1 × 16.00 g/mol = 28.01 g/mol
Fe
M(Fe) = 1 × M(Fe) = 1 × 55.85 g/mol = 55.85 g/mol
CO₂
M(CO₂) = 1 × M(C) + 2 × M(O) = 1 × 12.01 g/mol + 2 × 16.00 g/mol = 44.01 g/mol
Equation for half-lives:
Nt = No x (1/2)^n
No = initial amount
Nt = final amount after t years
n = number of half lives = t/(single half-life)
t = years
Nt = 3/12 = 0.25
No = 12/12 = 1.00
n = t/(24400)
3/12 = (12/12) x (0.5)^(t/24400)
(0.25) = 1.00 x (0.5)^(t/24400)
0.25/1.00 = 0.5^(t/24400)
ln(0.25) = ln(0.5^(t/24400))
ln(0.25) = (t/24400)*ln(0.5)
ln(0.25)/ln(0.5) = (t/24400)
2 = t/24400
2*24400 = t
t = 48800 yrs
answer is <u>t = 48,800 yrs</u>
Answer:
Explanation:
The answer is principle of original lateral
Continuity. Hope this helps
