A) saturated B)desaturated C)unsaturated D)supersaturated
1 answer:
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<h3>The molecular formula of this protein : C₁₈H₄₂O₁₂N₆</h3><h3>Further explanation </h3>
The empirical formula is the smallest comparison of atoms of compound forming elements.
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
The principle of determining empirical formula and molecular formula
- Determine the mass ratio of the constituent elements of the compound.
- Determine the mole ratio by dividing the elemental mass with the relative atomic mass obtained by the empirical formula
- Determine molecular formulas by looking for values of n
Find mol ratio for every component :
- C (Ar=12 g/mol):
- H(Ar=1 g/mol) :
- O(Ar=16 g/mol) :
N (r=14 g/mol) :
Mass of Nitrogen :
40.4-(17.16+3.17+13.71)=6.36 g
- N :
C : H : O : N = 1.43 : 3.17 : 0.857 : 0.454 = 3.15 : 7 : 1.89 : 1=3:7:2:1
Empirical formula : C₃H₇O₂N
Molecular mass of protein :
(C₃H₇O₂N)n=565.75
(12.3+1.7+2.16+14)n=565.75
(89)n=565.75
n=6.4≈6
so the molecular formula : C₁₈H₄₂O₁₂N₆
Answer:
- i) 5.00 mL of 1.00 M NaOH: before the equivalence point
- ii) 50.0 mL of 1.00 M NaOH: before the equivalence point
- iii) 100 mL of 1.00 M NaOH: at the equivalence point
- iv) 150 mL of 1.00 M NaOH: after the equivalence point
- v) 200 mL of 1.00 M NaOH: after the equivalence point
Explanation:
1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.
Equation:
- Molarity = numbrer of moles of solute / volume of the solution in liters.
Thus, you need to convert each volume from mL to liters, which is done dividing by 1,000.
Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:
2. Now calculate the number of moles of NaOH for every condition (addition)
<u>i) 5.00 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.
<u>ii) 50.0 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.
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<u>iii) 100 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.
<u>iv) 150 mL of 1.00 M NaOH</u>
Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.
<u>v) 200 mL of 1.00 M NaOH</u>
This is more volume of NaOH, then this is also after the equivalence point.