Answer:
c. 8.1 L
Explanation:
Given that:-
Moles of oxygen gas = 0.50 mol
According to the reaction shown below as:-
3 moles of oxygen gas on reaction gives 2 moles of ozone
Also,
1 mole of oxygen gas on reaction gives 2/3 moles of ozone
So,
0.50 mole of oxygen gas on reaction gives 
moles of ozone
Moles of ozone = 0.3333 mol
Pressure = 1 atm
Temperature = 25.0 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25.0 + 273.15) K = 298.15 K
Volume = ?
Using ideal gas equation as:
PV=nRT
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:
1 atm × V = 0.3333 mol × 0.0821 L.atm/K.mol × 298.15 K
⇒V = 8.1 L
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Answer:
C2H2O4
Explanation:
To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :
C = 26.86%
H = 2.239%
O = 100 - ( 26.86 + 2.239) = 70.901%
We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u
The division is as follows:
C = 26.86/12 = 2.2383
H = 2.239/1 = 2.239
O = 70.901/16 = 4.4313
We now divide each by the smallest number I.e 2.2383
C = 2.2383/2.2383 = 1
H = 2.239/2.2383 = 1
O = 4.4313/2.2383 = 1.98 = 2
Thus, the empirical formula is CHO2.
To get the molecular formula, we use the molar mass .
(CHO2)n = 90
We add the atomic masses multiplied by n.
(12 + 1 + 2(16))n = 90
45n = 90
n = 90/45 = 2.
Thus , the molecular formula is C2H2O4
This can be, for example, halogensubstituted hydrocarbons.
CCl₄, C₂F₆.
Or halides halocarboxylic acids, and other compounds.
O
II
Cl₃C-Cl
