Answer:
The molecular formula = 
Explanation:
Given that:
Mass of compound, m = 0.145 g
Temperature = 200 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (200 + 273.15) K = 473.15 K
V = 97.2 mL = 0.0972 L
Pressure = 0.74 atm
Considering,
Using ideal gas equation as:
where,
P is the pressure
V is the volume
m is the mass of the gas
M is the molar mass of the gas
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the values in the above equation as:-
The empirical formula is = 
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 12 + 1 = 13 g/mol
Molar mass = 78.31 g/mol
So,
Molecular mass = n × Empirical mass
78.31 = n × 13
⇒ n ≅ 6
The molecular formula =
THE KINETIC MOLECULAR THEORY STATES THAT ALL PARTICLES OF AN IDEAL GAS ARE IN CONSTANT MOTION AND EXHIBITS PERFECT ELASTIC COLLISIONS.
Explanation:
An ideal gas is an imaginary gas whose behavior perfectly fits all the assumptions of the kinetic-molecular theory. In reality, gases are not ideal, but are very close to being so under most everyday conditions.
The kinetic-molecular theory as it applies to gases has five basic assumptions.
- Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size.
- Gas particles are in constant rapid motion in random directions.
- Collisions between gas particles and between particles and the container walls are elastic collisions.
- The average kinetic energy of gas particles is dependent upon the temperature of the gas.
- There are no forces of attraction or repulsion between gas particles.
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
______
NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
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Hope this helps guys!
