Answer:I think it’s mechanical energy
Explanation:
Answer is: 8,7g.
Chemical reaction: 2Al + 6HCl → 2AlCl₃ + 3H₂.
m(Al) = 78,4g.
n(Al) = m(Al)÷M(Al) = 78,4g ÷ 27g/mol = 2,9 mol.
n - amount of substance.
from reaction: n(Al) : n(H₂) = 2 : 3.
2,9mol : n(H₂) = 2 : 3.
n(H₂) = 4,35 mol.
m(H₂) = n(H₂)·M(H₂) = 4,35mol · 2g/mol = 8,7g.
Well, you don't need enzymes (biological catalysts) if you're willing to wait a century or two to digest a burger.
Without catalysts, complex reactions like digestion would take too long and the organism could not extract energy from the nutrients it eats in a practical time frame.
In addition, speed is everything in the biological world.
Some reactions and their speed relative to other organisms reactions determines who survives and who doesn't, among other aspects of life.
If a plant is slow to photosynthesize and grow in a habitat high in competition for sunlight real estate, other autotrophs will surely take over.
Answer:
Chemotherapy drugs and other toxins are actively pumped out of cancer cells by transmembrane proteins.
Explanation:
Drugs inside the cell can be inactivated by oxidation and / or conjugation with glutathione, such as glutathione S-transferases, playing an important role in detoxification. However, conjugation is not enough for drug elimination. And this is where the GS-X pumps appear. Transporter proteins, such as Mrp protein, act as GS-X pumps. The PgP, Mrp and Bcrp proteins function as expulsion pumps, thus reducing the intracellular accumulation of drugs, causing resistance in cancer cells.
First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
