Answer:
B
101L
Explanation:
We use the ideal gas relation
PV = nRT
P = pressure = 101.3KPa
V = volume = ?
n = number of moles = 4.5moles
T = Temperature = 273.15K
R = molar gas constant = 8.314J/mol.k
Rearranging the equation to make V the subject of the formula yields :
V = nRT/P
= ( 4.5 × 8.314 × 273.15) ÷ 101.3
= 10,219.361 ÷ 101.3 = 100.88L which is apprx 101L
Answer:
For the control experiment: of aqueous solution of erioglaucine has absorbance of
From Lambert-Beer's law we know:
Here; e is the molar absorptivity coefficient of erioglaucine
l = length of cuvette in which the solution is taken =
A sorbance by the erioglaucine = total absorbance - absorbance by distilled
So; by putting the values in the above equation; we get:
So;
The molar absorptivity coefficient of erioglaucine is
The absorbance of erioglaucine in distilled water (contaminated with metal ions) is:
The absorbance of distilled water is
So; absorbance of erioglaucine itself is :
Again using Lambert Beer law; we get:
c = 0.198/5.65 M = 0.035 M
The concentration of the erioglaucine is
Explanation:
Answer:
Answer: The mass of product left in the test tube will be less than the initial measured mass of the reactants.
Explanation: This is because one of the products of the reaction is a gas (hydrogen) and it escapes as the reaction happens in an open system.
This is the reaction,
Mg(s) + 2HCl(aq) --------> MgCl2(aq) + H2(g)
So, evidently, only MgCl2 is left in the reaction test tube together with unreacted reactants.
But, the hydrogen gas that escapes accounts for the lesser mass at the end of the reaction.
The mass in the test tube at the end of the reaction +
the mass of Hydrogen gas that escapes = the mass of reactants before the reaction; consolidating the law of conservation of mass.
Answer:
7.28 × 10³ K
Explanation:
Let's consider the following reaction.
N₂(g) + O₂(g) → 2 NO(g)
The reaction is spontaneous when the standard Gibbs free energy (ΔG°) is negative. ΔG° is related to the standard enthalpy of the reaction (ΔH°) and the standard entropy of the reaction (ΔS°) through the following expression.
ΔG° = ΔH° - T . ΔS°
If ΔG° < 0,
ΔH° - T . ΔS° < 0
ΔH° < T . ΔS°
T > ΔH°/ΔS° = (180.5 × 10³ J/mol)/(24.8 J/mol.K) = 7.28 × 10³ K
The reaction is spontaneous above 7.28 × 10³ K.