Answer:
31.60% phosphorus
Explanation:
To find the percent by mass, you first calculate the total mass and then the mass of phosphorus, and finally divide phosphorus mass by total.
Phosphorus mass: the molar mass of P is 30.97 g/mol, and since there's 1 mole of P, we just have 30.97 g.
Total mass: we add all the molar masses of the components together.
We have 3 moles of H, so we multiply 3 by 1.008 g/mol = 3.024 g H.
We already calculated the mass of phosphorus: 30.97 g P.
We have 4 moles of O, so we multiply 4 by 16.00 g/mol = 64.00 g O.
The total is then the sum: 3.024 + 30.97 + 64.00 = 97.994 g ≈ 97.99 g
Now, to find the percentage, we take 30.97 g P and divide by 97.99:
30.97/97.99 ≈ 0.3160 ⇒ 31.60% P
Thus, the answer is 31.60% phosphorus.
Hope this helps!
Divide the mass of the proton by the mass of the electron.
Answer:
See explanation
Explanation:
In order to do this, we need to use 3 reagents to get the final product.
The first one, and logic is the halogenation of the alkene. Doing this, with Br2/CCl4, we'll get an alkane with two bromines, one in carbon 2 and the other in carbon 3.
Then, the next step is to eliminate one bromine of the reactant. The best way to do this, is using sodium ethoxide in ethanol. This is because sodium ethoxide is a relatively strong base, and it will promove the product of elimination in major proportions rather than the sustitution product. If we use NaOH is a really strong base, and it will form another product.
When the sodium ethoxide react, it will form a double bond between carbon 1 and 2 (The carbon where one bromine was with the methyl, changes priority and it's now carbon 3).
The final step, is now use acid medium, such H3O+/H2O or H2SO4/H2O. You can use any of them. This will form an carbocation in carbon 2 (it's a secondary carbocation, so it's more stable that in carbon 1), and then, the water molecule will add to this carbon to form the alcohol.
See the attached picture for the mechanism of this.
2-Methyl-4-oxo-pentanoic acid is unlikely to produce 2-Methyl-3-butanone upon strong heating.
Upon heating, the β ketoacid becomes unstable and decarboxylates, leading to the formation of the methyl ketone.
A carboxylic acid is an organic acid that contains a carboxyl group (C(=O)OH) attached to an R-group. The general formula of a carboxylic acid is R−COOH or R−CO2H, with R referring to the alkyl, alkenyl, aryl, or other group.
Carboxylic acids occur widely. Important examples include the amino acids and fatty acids. Deprotonation of a carboxylic acid gives a carboxylate anion.
Full question :
Q. Which reactant is unlikely to produce the indicated product upon strong heating?
- A) 2,2-Dimethylpropanedioic acid 2-methylpropanoic acid
- B) 2-Ethylpropanedioic acid Butanoic acid
- C) 2-Methyl-3-oxo-pentanoic acid 3-Pentanone
- D) 2-Methyl-4-oxo-pentanoic acid 2-Methyl-3-butanone
- E) 4-Methyl-3-oxo-heptanoic acid 3-Methyl-2-hexanone
Hence, option (D) is correct.
Learn more about carboxylic acid here : brainly.com/question/26855500
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