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2 years ago

Why are carboxylic acids more acidic than alcohols or alkynes? (choose all that apply)

Chemistry

1 answer:

wlad13 [49]2 years ago

6 0

Answer:

The answers are (A.) Carboxylate ions have resonance, while alkoxides and acetylides don't (D.) Both carboxylic acids and carboxylates have resonance stabilization

Explanation:

When alcohols or alkynes donate a proton (H+), they become alkoxides acetylides respectively. Carboxylic acids on the other hand form carboxylate ions.

Carboxylate ions are more stable than their alkoxide and acetylides counterparts, because they possess a resonance structure which disperses its negative charge and allows for the valence electrons to be effectively delocalized.

Hence, carboxylic are more acidic than alcohols or alkynes.

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As part of a science experiment, Jose did a test for starch on a slice of apple and a slice of potato. The yellow-orange iodine

juin [17]

Answer:

See explanation

Explanation:

Iodine solution is used to test for starch. A positive test for starch gives a blue-black color.

The fact that the color of the apple remained the same is indicative of the fact that starch was not contained in the apple.

A change in the color of potato indicates the presence of starch in the potato.

The fact that iodine did not react with apple should not be taken to mean that apples contain no starch at all. Starch changes gradually to sugar as fruits ripen. This is why the apple gave a negative test for starch.

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3 years ago

If an atom has no electric charge, what can be said about the number of protons and electrons it contains?

vitfil [10]

then the electrons and protons would have a even amount of negetive electric charges

8 0

2 years ago

Please help with word equations?

Delvig [45]

For the answer to the questions above,

a) Ag2CO3(s) => Ag2O(s)+CO2(g)

b) Cl2(g)+2(KI)(aq) => I2(s)+2(KCl)(aq) (coefficients are for balanced equation) 
net ionic is Cl2(g)+2I- => I2(s)+2Cl-(aq) 

c) I2(s)+3(Cl2)(g)=>2(ICl3)
I hope I helped you with your problem

7 0

3 years ago

What is the cell potential for the reaction mg(s+fe2+(aq?mg2+(aq+fe(s at 77 ?c when [fe2+]= 3.40 m and [mg2+]= 0.210 m . express

Galina-37 [17]

First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential

Fe+2 + 2e- --> Fe potential= -0.44
Mg+2 + 2e- --> Mg potential= -2.37

Cell potential= (-0.44) + (+2.37)= 1.93 V

Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.

Download pdf

8 0

2 years ago

You are given a solid that is a mixture of na2so4 and k2so4.

Murljashka [212]

Here we have to calculate the amount of $SO_{4}^{2-}$ ion present in the sample.

In the sample solution 0.122g of $SO_{4}^{2-}$ ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ + $SO_{4}^{2-}$

(Na)₂SO₄=2Na⁺ + $SO_{4}^{2-}$

Thus, BaCl₂+ $SO_{4}^{2-}$ = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of $SO_{4}^{2-}$ ion is precipitated in this reaction.

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion ( $SO_{4}^{2-}$ ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present. So in 1 g of BaSO₄ $\frac{96.06}{233.3}=0.411$ g of $SO_{4}^{2-}$ ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of $SO_{4}^{2-}$ ion is present.

5 0

3 years ago