Answer:
0.861 L
Explanation:
We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:
(P₁V₁)/T₁ = (P₂V₂)/T₂
Convert the temperatures to degrees Kelvin.
25.0°C -> 298 K, 100.0°C -> 373 K
Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:
(165(2.5))/298 = (600(V₂))/373
V₂ = (165(2.5)(373))/(298(600))
V₂ = 0.861 L
MASS: generally a measure of an object's resistance to changing its state of motion when a force is applied. It is determined by the strength of its mutual gravitational attraction.
Molecular mass or molecular weight is the mass of a molecule.
Gram formula mass (a.k.a. molar mass) is defined as the atomic mass of one mole of an element, molecular compound or ionic compound. The answer must always be written with the unit g/mol (grams per mole).
Answer:
Using the information provided in the question, the percentage by mass of ethanol in the blood is 0.1949%.
Explanation:
The balanced equation for the reaction can be written as:
3CH3CH2OH + 2K2Cr2O7+ 16H+ >>>> 3CH3COOH + 4Cr3+ + 4K+ + 11H2O. This will be used to solve the problem.
The no of moles of K2Cr2O7 used = concentration X volume in dm cube.
= 0.05 X 22.6/1000 = 1.13/1000 = 0.00113 moles of K2Cr2O7 was used.
From the balanced equation above, 3 moles of ethanol will react with 2 moles of K2Cr2O7.
Therefore, 0.00113 X 3/2 moles of ethanol would have been in the blood of the subject = 0.001695 moles of ethanol.
Remember that no of moles = mass/molar mass. Then mass = moles X molar mass.
The molar mass of ethanol is 46g/mole.
Then mass = 0.001695 X 46 = 0.07797gram of Ethanol was in the blood.
Percentage by mass in blood is: 0.07797/40 X 100 = 0.1949%
The mole ratio of PbO2 to water in
Pb + PbO2 +2 H2SO4 = 2PbSO4 + 2H2O is
1 :2
1 mole of Pb reacted with 1 mole of PbO2 and 2 moles of H2SO4 to form 2 moles of PbSO4 and 2 moles of H2O
therefore the mole ratio of Pbo2 to H2O is 1:2
Looks like exponential decay. Collision theory is consistent with this model because at higher reactant concentrations we expect the reactants to encounter each other more often (collide) and as the concentration decreases we expect fewer and fewer collision events so the rate of reaction becomes less and less.
