Answer: Option (3) is the correct answer.
Explanation:
Aerobic organisms are the organisms which survive and grow in the presence of oxygen.
When oxidation of glucose occurs in the presence of oxygen then it is known as aerobic respiration.
In aerobic respiration, food releases energy to produce ATP which is necessary for cell activity. There is complete breakdown of glucose in aerobic respiration that is why more energy is released. Therefore, aerobic organisms become active.
Thus, we can conclude that characteristics very active, efficient use of energy describes aerobic organisms.
Answer:
The solution is given below
Explanation:
Heat, q= mc∆T
q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C
q= -1619.75J
NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.
Enthalpy Change, ∆H = 1619.75 7/ 10.5 g
= 154.26 J/g
No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr
=10.5g/119gmol-1
=0.088 mol
∆H= 1619.75 J/ 0.088 mol
= 18.41 kJ/mol
Answer:
1. The correct option is;
c. maintains charge balance in the cell
2. The correct option is;
c. +3.272 V
Explanation:
The aqueous solution in a galvanic cell is the electrolyte which is a ionic solution containing that permits the transfer of ions between the separated compartment of the galvanic cell such that the overall system is electrically neutral
Therefore, the aqueous solution maintains the charge balance in the cell
2. Here we have;
B₂ + 2e⁻ → 2B⁻ Ecell = 0.662 V
A⁺ + 1e⁻ → A Ecell = -1.305 V
Hence for the overall reaction, we have;
2A + B₂ → 2AB gives;
(0.662) - 2×(-1.305) = +3.272 V.
Answer:
c.
Explanation:
A serial dilution is a dilution that is made fractionated. The stock solution is diluted, then this now solution is diluted, and then successively. The final dilution is the multiplication of the steps dilutions.
The representation of the dilution is v/v (volume per volume) indicates how much of the stock solution is in the total volume of the solution. So 1/5 indicates 1 mL to 5 mL of the solution. If the final volume must be 1 mL, then the stock solution must be 0.2 mL (0.2/1 = 1/5), and the volume of the solvent is 1 mL - 0.2 = 0.8 mL.
The second solution is done with a dilution of 1/10 or 1 mL of the first solution in 10 mL of the total solution. Because the solution has 1 mL, then the volume of the first solution must be 0.1 mL (0.1/1 = 1/10), and the volume of the solvent that must be added is 1 mL - 0.1 mL = 0.9 mL.
