There's a lot of capillaries in the lungs because the blood needs to be transferred and the capillaries are the smallest vessels that can do this.
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A 25.00 ml sample of hydrochloric acid solution, HCl, is titrated with 0.0512 m NaOH solution. the volume of NaOH solution required is 21.68 ml then the molarity of the HCl solution is 0.044 M .
Calculation ,
Formula used : 
...( i )
Where M is the molarity or concentration and V is the volume in ml .
concentration of hydrochloric acid solution ( 
) = ?
concentration of NaOH ( 
) = 0.0512 M
volume of hydrochloric acid solution ( 
) = 25.00 ml
volume of NaOH ( 
) = 21.68 ml
Putting the value of concentration , volume of both in equation ( i ) we get .
× 25.00 ml = 0.0512 × 21.68 ml
= 0.0512 × 21.68 ml / 25.00 ml= 0.044 M
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Answer:
A.) trans-2,3-dichloro-5-methyl-2-hexene
Explanation:
The proposed question and structure has been attached to this answer.
The<em> cis</em>- and <em>trans</em>- isomers are used to denote the alignment of the functional groups to the carbon chain side. From the diagram, we see that:
- the two chloro groups on the main compound are in opposite directions (-<em>trans</em>), and not the same sides (-<em>cis</em>)
- The bottom extended methyl group is on the 5th carbon as we count from left to right
- there's a total of 6 carbon atoms with 1 double bond on the second to third carbon.
Hence the name <u>trans-2,3-dichloro-5-methyl-2-hexene</u>
For the first compound,
Molar Mass of C=12.01x 2 =24.02
Molar Mass of H= 1.01x 6= 6.06
So molar mas of C2H6= (12.01x2)+(1.01x6) = 30.08
To get the percent composition, the formula is “(Total MM of C in molecule)/(Total MM of molecule)x 100”.
Here is the solution:
% of C= 24.02/30.08 x 100 = 79.9
% of H=6.06/30.08 x 100=20.1
To check if it is equal to 100%, you just have to add the percentages.
For the second compound,
Molar Mass of Na=22.99 x 1 =22.99
Molar Mass of H= 1.01x 1= 1.01
Molar Mass of S= 32.07x 1= 32.07
Molar Mass of O = 16.00 x4 = 64.00
So molar mas of NaHSO4= (22.99)+(1.01)+(32.07)+(16.00x4) = 120.07
To get the percent composition, the formula is “(Total MM of C in
molecule)/(Total MM of molecule)x 100”.
Here is the solution:
% of Na= 22.99/120.07 x 100 = 19.2
% of H= 1.01/120.07x 100 = 0.8
% of S= 32.07/120.07x 100 = 26.7
% of O= 64/120.07 x 100 = 53.3
To check if it is equal to 100%, you just have to add the percentages.
