Answer:
the carbohydrates in cotton are mainly cellulose, but our body does not possess the proper enzymes to process cellulose, no we don't eat cotton on in our daily diet.
Hey there!:
V1 = 3.05 L
V2 = 3.00 L
P1 = 724 mmHg
P2 = to be calculated
T1 = 298 K
T2 = 273 K
Therefore:
P1*V1 / T1 = P2*V2 / T2
P2 = ( P1*V1 / T1 ) * T2 / V2
P2 = 724 * 3.05 * 273 / 298 * 3.00
P2 = 602838.6 / 894
P2 = 674.31 mmHg
1 atm ----------- 760 mmHg
atm ------------- 674.31 mHg
= 674.31 * 1 / 760
= 0.887 atm
Hope this helps!
London dispersion is present, as we can see C3H5(OH)3 is very branched, when a molecule is branched it has a low boiling point due to how compact the surface area is compare to C3H7OH which has a large surface area (in this molecule there is only on branch)
First, we have to get the initial [C6H8O6] = mass/molar mass
when the molar mass of C6H8O6 = 176.12 g/mol
∴[C6H8O6] = 0.25 g / 176.12 g/mol
= 0.00142 M
when
C6H8O6 ⇄ H+ + C6H7O6-
intial 0.00142 M 0 0
change -X +X +X
Equ (0.00142-X) X X
so, Ka = [H+][C6H7O6-] / [C6H8O6]
by substitution:
8 x 10^-5 = X * X / (0.00142-X) by solving this equation for X
∴ X = 0.000299
∴[H+] = 0.000299
∴PH = -㏒[H+]
= -㏒ 0.000299
= 3.52
