Calculate the mole of glucose and water
The moles of water =158g/18g/mol=8.778 moles
moles of glucose =52.8g/180g/mol=0.293 moles
determine the mole fraction of the solvent
that is x solvent = 8.778/ (8.778+0.293)=0.9677
use the Raults law to determine the vapor pressure
100 degree of water has a vapor pressure of 760 mmhg
p solution=(x solvent) (p^o solvent)
p solution=0.9677 x760 =735.45 mmHg
<span>The law of proportion states that elements combine in whole number ratios. The gram readings for K are multiples of each other, both in grams and moles.
Let us compare the ratios:
</span>2.44 grams/1.22 grams = 2
<span>4.89 grams/2.44 grams = 2</span>
<span>Therefore, Potassium always combines with Oxygen in a ratio of 2 is to 1.</span>
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 total pressure is 1.107 atm.
Explanation;
The total pressure is the sum of the pressures of the three gases in the flask
Pressure (total) = 0.215 atm + 0.066 atm + 0.826 atm = 1.107 atm
= 1.107 atm.
