Answer:
Give them each one so all of you is 8
Explanation:
I hope it helps:)
Answer:
Explanation:
Diethyl malonate is more acidic than monocarbonyl compounds (pKa=13) due to their alpha hydrogens being attached to two carbonyl groups. Thus, the <u>malonic ester is easily converted to its enolate ion by reaction with sodium ethoxide in ethanol</u>. The product of the alkylation of the malonic ester leaves a <u>hydrogen atom acid in an alpha position</u>, so the alkylation process can be repeated a second time to produce a dialkylated malonic ester.
In this case, when urea is treated with diethyl malonate in the presence of sodium ethoxide base,<u> the second alkylation step occurs intramolecularly</u> to generate a cyclic product, barbituric acid.
Well you did not show the figure, but I can show you how to do it.
Suppose the person is 6 feet tall and the bacteria cell is 1 micrometer in length.
So, put both magnitudes in the same units. I will use meter.
Person: 6 feet * 0.3048 m / foot = 1.8288 m, which in orders of magnitud is about 2 m
Bacteria cell: 1 micrometer * 1 m / 1,000,000 micrometer = 1 * 10 ^ -6 m.
Now divide: 2m / 10^ -6 m = 2,000,000
So a person is about 2 million times larger than a bacteria cell.
If the numbers in your figure are different you can use the same method to get a new result according to your figure.
Answer:
B. CH3COOH pH > 4.7 (4.8)
Explanation:
- CH3COOH + NaOH ↔ CH3COONa + H2O
- CH3COONa + NaOH ↔ CH3COONa
∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol
⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)
⇒ <em>C</em> CH3COOH = 0.0905 M
∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol
⇒ <em>C</em> CH3COONa = (0.01 mol + 5 E-4 mol) / (0.105 L )
⇒ <em>C</em> CH3COONa = 0.1 M
∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5
⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])
⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]
⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0
⇒ [H3O+] = 1.5835 E-5 M
∴ pH = - Log [H3O+]
⇒ pH = - Log (1.5835 E-5)
⇒ pH = 4.8004 > 4.7
Answer:
CH3—CH2—O—CH2—CH3 is the compound of glycol.