Answer:
Each specific property of motif and domain is explained.
Explanation:
Domain;
- May retain a 3D structure when separated from rest of the protein.
- Unit of tertiary structure because alpha helix and beta sheets are units of secondary structure.
- Stable globular units like pyruvate kinase
- May be distinct functional units in a protein
Motif;
- Repetetive supersecondary structure because they contain cluster of secondary structure.
- Beta Alpha Beta unit is an example of motif
- Clusters of secondary structure
Both Motif and Domain;
- Stabilized by hydrophobic interactions like hydrogen bonding stabilize the both.
- Depends on primary structure like the arrangement of amino acid in polypeptide chain determine the secondary and tertiary structure of proteins.
Answer: a) 
b) 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
a) Mass of Ba= 66.06 g
Mass of Cl = 34.0 g
Step 1 : convert given masses into moles.
Moles of Ba =
Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Ba =
For O =
The ratio of Ba: Cl= 1:2
Hence the empirical formula is 
b) Mass of Bi= 80.38 g
Mass of O= 18.46 g
Mass of H = 1.16 g
Step 1 : convert given masses into moles.
Moles of Bi =
Moles of O=
Moles of H=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Bi=
For O =
For H=
The ratio of Bi: O: H= 1:3: 3
Hence the empirical formula is 
Answer:
1. Gases can be easily liquefied into very small volumes and stored in liquid form Eg in LPGA cylinders and used in homes.
2. Balloons can be easily filled with air.
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
The correct answer is B. H2SO4 + B(OH)3 B2(SO4)3 + H2O
Hope this helps!