D.10.0 mol is the correct
Answer:
Explanation:
Given:
Pressure = 745 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 745 / 760 = 0.9803 atm
Temperature = 19 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (19 + 273.15) K = 292.15 K
Volume = 0.200 L
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K
⇒n = 0.008174 moles
From the reaction shown below:-
1 mole of react with 2 moles of
0.008174 mole of react with 2*0.008174 moles of
Moles of = 0.016348 moles
Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)
So,
Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).
Answer:
C₃H₆O₃
Explanation:
Data:
EF = CH₂O
MM = 90. g/mol
Calculations:
EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ
MF = (CH₂O)₃ = C₃H₆O₃
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj