The concentration of acetic acid in the vinegar is found to be 0.5857 M.
<u>Explanation:</u>
Here one mole of acetic acid is used to neutralize one mole of sodium hydroxide. According to the law of Volumetric analysis, we can write the equation as,
V1M1 = V2M2
V1 and M1 being the volume and molarity of NaOH
V2 and M2 being the volume and molarity of CH₃COOH
M2 = 
= 
= 0.5857 M
So the concentration of acetic acid in the vinegar is 0.5857 M.
Answer:
Please find the structure attached as an image
Explanation:
Based on the characteristics ending name (-ene) of the organic compound above, it belongs to the ALKENE GROUP. Alkenes are characterized by the possession of a carbon to carbon double bond (C=C) in their structure.
- But-3-ene tells us that the organic compound has four straight carbon atoms with the C=C (double bond) located on the THIRD carbon depending on if we count from right to left or vice versa.
- 2 methyl indicates that the methyl group (-CH3) is located as an attachment on the second carbon (carbon 2).
N.B: In the structure attached below, the counting is from the left to right (→).
Because a unexpected color change has taken place also a precipitate was formed
Answer:
The answer is B
Explanation:The answer is B because the color would be irreversible.
Answer:
6.69%
Explanation:
Given that:
Mass of the fertilizer = 0.568 g
The mass of HCl used in titration (45.2 mL of 0.192 M)
= 
= 0.313 g HCl
The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)
= 
= 0.0058919 mole of NaOH
From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process
Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g
= 0.2151 g HCl
From above ; the total amount of HCl used = 0.313 g
The total amount that is used for complete neutralization = 0.2151 g
∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g
= 0.0979 g
We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g
Now; the amount of HCl neutralized by 0.0979 HCl = 
= 0.0456 g
Therefore, the mass of nitrogen present in the fertilizer is:
= 
= 0.038 g
∴ Mass percentage of Nitrogen in the fertilizer =
%
= 6.69%