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Nimfa-mama [501]
3 years ago
14

Draw the isomers of a compound with the molecule formula C4H4O4. Also state the isomerism exhibited by them.

Chemistry
1 answer:
Maslowich3 years ago
5 0

cis-2- butene-1,4-dioic acid and trans-2- butene-1,4-dioic acid are the 2 isomers with the molecular formula C₄H₄O₄.

<u>Explanation:</u>

Isomers are the structures having the same molecular formula but their is a difference in their structural formula and this concept is known as isomerism.

Here we have the molecular formula C₄H₄O₄ and it may be dicarboxylic acid and it may be exist as geometrical isomers as cis-2- butene-1,4-dioic acid and trans-2- butene-1,4-dioic acid.

The isomers are given.

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aluminum bronze contains 92.0% of copper and 8.0% aluminum. what maximum mass of aluminum bronze can be prepared from 73.5g of c
slega [8]
The answer is 79.9 g.

Copper takes 92.0% of aluminum bronze and it is a limiting factor. We have aluminum in excess, so we need to make a proportion.
If 73.5 g of copper are 92.0%, how many g of aluminum bronze will be 100%:
73.5 g : 92.0% = x : 100%.
x = 73.5 g : 100% * 92.0%
x = 79.9 g

Therefore, from 73.5 g of copper and 6.4 g of aluminum (since 79.9 g - 73.5 g = 6.4 g), maximum 79.9 g of aluminum bronze can be prepared.
4 0
3 years ago
A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to
melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:

  • pH = pKa + log\frac{[CH_3COO^-]}{[CH_3COOH]}

For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

  • CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
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The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • pH = 4.75 + log\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L} = 5.54
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3 years ago
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