Answer:
24e⁻ are transferred by the reaction of respiration.
Explanation:
C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂
This is the reaction for the respiration process.
In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION
Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION
6C → 6C⁴⁺ + 24e⁻
In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻
6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²
In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.
Answer:
The open system evaporates the solvent in the solution
Explanation:
An open system is a system in which exchange of materials and energy can occur. If a TLC set up is left open, then the set up constitutes an open system.
During TLC, the sample is dotted on the plate and inserted into a suitable solvent. The solvent moves up the plate and achieves the required separation of the mixture.
Most of these solvents used used TLC are volatile organic compounds. Therefore, if the TLC set up is left open, the solvent will evaporate leading to poor results after running the TLC experiment.
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
<span>1.05 g/ml * 1000 ml = 1050g/l because of 1g/ml = 1 kg/l
so, a/q
mass of 4.7 l of whole blood in pound =
4.7 * 1050 = 4935 g
so in pound
4935g = 10.87981p</span>
Answer:
Approximately 0.126 M
Explanation:
For the calculation of the dilution you take into account the moles of NaOH in the 42.1mL of the original solution and you use the new volume of 342.1 mL:
The standardization is necessary because a beaker is not not an instrument used to measure volumes and the marks on it only give an estimate of the volume of the solution, they are used to contain solutions and carry reactions among other things. If you would have measured the water with a graduated cylinder (an instrument designed to measure volumes) the standardization wouldnt be that necessary.
