Answer:
4
General Formulas and Concepts:
<u>Chemistry</u>
- Identifying Moles/Atoms and Compounds
Explanation:
<u>Step 1: Define</u>
Al₂(SO₃)₃ - Aluminum Sulfide
2Al₂(SO₃)₃
<u>Step 2: Identify</u>
In Aluminum Sulfide, we have 2 Al's for every 3 SO₃²⁻'s.
So far, we have 2 Al's in total.
BUT, since we have a 2 coefficient in 2Al₂(SO₃)₃, we need to multiply by 2.
Therefore, we have 4 Al's in total.
<h3>The molecular formula of this protein : C₁₈H₄₂O₁₂N₆</h3><h3>Further explanation </h3>
The empirical formula is the smallest comparison of atoms of compound forming elements.
A molecular formula is a formula that shows the number of atomic elements that make up a compound.
(empirical formula) n = molecular formula
The principle of determining empirical formula and molecular formula
- Determine the mass ratio of the constituent elements of the compound.
- Determine the mole ratio by dividing the elemental mass with the relative atomic mass obtained by the empirical formula
- Determine molecular formulas by looking for values of n
Find mol ratio for every component :
![\tt \dfrac{17.16}{12}=1.43](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B17.16%7D%7B12%7D%3D1.43)
![\tt \dfrac{3.17}{1}=3.17](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B3.17%7D%7B1%7D%3D3.17)
![\tt \dfrac{13.71}{16}=0.857](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B13.71%7D%7B16%7D%3D0.857)
N (r=14 g/mol) :
Mass of Nitrogen :
40.4-(17.16+3.17+13.71)=6.36 g
![\tt \dfrac{6.36}{14}=0.454](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B6.36%7D%7B14%7D%3D0.454)
C : H : O : N = 1.43 : 3.17 : 0.857 : 0.454 = 3.15 : 7 : 1.89 : 1=3:7:2:1
Empirical formula : C₃H₇O₂N
Molecular mass of protein :
![\tt \dfrac{40.4}{0.07141}=565.75](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B40.4%7D%7B0.07141%7D%3D565.75)
(C₃H₇O₂N)n=565.75
(12.3+1.7+2.16+14)n=565.75
(89)n=565.75
n=6.4≈6
so the molecular formula : C₁₈H₄₂O₁₂N₆
Answer:
The pH to be 10 cm from the most acidic end is 3.42.
Explanation:
The pH at one end = 1
The pH at another end = 13
Length of the chamber = 13 cm
Change in pH with respect to length of the chamber from acidic end = ![x=\frac{13-1}{35 cm}=\frac{12}{35} pH/cm](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B13-1%7D%7B35%20cm%7D%3D%5Cfrac%7B12%7D%7B35%7D%20pH%2Fcm)
So, the pH of the chamber 10 cm from the most acidic end:
![\frac{12}{35} pH/cm\times 10 cm = 3.42](https://tex.z-dn.net/?f=%5Cfrac%7B12%7D%7B35%7D%20pH%2Fcm%5Ctimes%2010%20cm%20%3D%203.42)
The pH to be 10 cm from the most acidic end is 3.42.