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neonofarm [45]
2 years ago
14

When dissolved in water, all acids will

Chemistry
2 answers:
yarga [219]2 years ago
8 0
B is the answer I think but I hope it is
poizon [28]2 years ago
7 0
When dissolved in water, acids donate hydrogen ions (H+). Hydrogen ions are hydrogen atoms that have lost an electron and now have just a proton, giving them a positive electrical charge. ... If a solution has a high concentration of H+ ions, then it is acidic.
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Describe the contributions of 3 scientists to our current understanding of the atom
bogdanovich [222]

Answer:

1. Rutherford did the scattering experiment and observed that some of the rays bounce back. He concluded that there is a mass in which positive charge is concentrated. This marks the discovery of nucleus.

2. J.J Thomson discovered electrons by conducting cathode ray experiment.

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8 0
3 years ago
Based on the collision theory, which factors are likely to increase the rate of reaction? Select all that apply.
Alexus [3.1K]

Answer:

Option a

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7 0
2 years ago
Draw the product that valine forms when it reacts with excess CH3CH2OH and HCl followed by a wash with aqueous base.
-BARSIC- [3]

Answer:

Product: ethyl L-valinate

Explanation:

If we want to understand what it is the molecule produced we have to an<u>alyze the reagents</u>. We have valine an <u>amino acid</u>, in this kind of compounds we have an <em>amine group</em> (NH_2) and a <em>carboxylic acid</em> group (COOH).  Additionally, we have an <u>alcohol </u>(CH_3CH_2OH) in the presence of HCl (a <u>strong acid</u>) in the first step, and a base (OH^-).

When we have an acid and an alcohol in a vessel we will have an <u>esterification reaction</u>. In other words, an ester is produced. As the <em>first step,</em> the oxygen in the C=O (in the carboxylic acid group) would be protonated. In the <em>second step</em>, the ethanol attacks the carbon in the C=O of the carboxylic acid group producing a new bond between the oxygen in the ethanol and the carbon in the carboxylic acid. In <em>step 3</em>, a proton is transferred to produce a better leaving group (H_2O). In <em>step 4</em>, a water molecule leaves the main structure to produce again the double bond C=O. <em>Finally</em>, a base (OH^-) removes the hydrogen from the C=O bond to produce ethyl L-valinate

See figure 1

I hope it helps!

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3 years ago
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(HELP ME ASAP!)
slega [8]
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8 0
3 years ago
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