Answer:
C₅H₁₀O₅
Explanation:
1. Calculate the mass of each element in 2.78 mg of X.
(a) Mass of C

(b) Mass of H

(c) Mass of O
Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g
2. Calculate the moles of each element

3. Calculate the molar ratios
Divide all moles by the smallest number of moles.

4. Round the ratios to the nearest integer
C:H:O = 1:2:1
5. Write the empirical formula
The empirical formula is CH₂O.
6. Calculate the molecular formula.
EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ

MF = (CH₂O)₅ = C₅H₁₀O₅
The molecular formula of X is C₅H₁₀O₅.
The IUPAC name for the given product is 2 chloro Butane.
<h3>What is IUPAC nomenclature?</h3>
IUPAC stands for 'International Union of Pure and Applied Chemistry', which givers some rule for designing the name of compounds of chemistry.
- In the given product total four carbon atoms are present and between all of them single bonds are present.
- In the second carbon atom, chlorine group is present.
- During the nomenclature process, first we write down the name of the attached group which is followed by the alkane chain.
Hence name of the product is 2 chloro Butane.
To know more about IUPAC nomenclature, visit the below link:
brainly.com/question/26635784
#SP1
Answer:
When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.
Aldolase cleaves a hexose into two trioses.
[See the image attached].
Asterisk indicates the label.
When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.
<span>Mass of the solution = 0.17m
Kb for C6H5NH2 = 3.8 x 10^-10
We know Ka for C6H5NH2 = 1.78x10^-11
We have Kw = Ka x Kb => Ka = Kw / Kb
=> (C2H5NH2)(H3O^+)/(C2H5NH3^+) => 1.78x10^-11 = K^2 / 0.17
K^2 = 3 x 10^-12 => K = 1.73 x 10^-6.
pH = -log(Kw(H3O^+)) = -log(1.73 x 10^-6) = 5.76</span>