Answer: False. Hope this helps!!
Explanation:
The given compounds are oxyacids and in these compounds more is the electronegativity of the central atom more will be its acidic strength.
This is because more is the electronegativity of the central atom more will be the polarity of OH bond. As a result, the compound can readily lose 
ion.
Also, more is the electronegativity of central atom more will be the stability of conjugate base formed.
Thus, we can conclude that given compounds are arranged in increasing acid strength as follows.
HOI < 
< 
< HOF
41.083 atm is the difference between the ideal pressure (as predicted by the ideal gas law) and the real pressure (as predicted by the van der Waals equation.
Explanation:
Data given for argon gas:
number of moles = 1 mole
volume = 0.5 L
Temperature = 19 degrees or 292.15 K
a= 1.345 (L2⋅atm)/mol2
b= 0.03219L/mol.
R = 0.0821
The real pressure equation given by Van der Waals equation:
P =( RT ÷ Vm-b) - a ÷ Vm^2
Putting the values in the equation:
P = (0.0821 x 292.15) ÷(0.5 - 0.03219) - 1.345÷ (0.5)^2
= 23.98÷0.4678 - 1.345 ÷0 .25
= 51.26 - 5.38
= 45.88 atm is the real pressure.
The pressure from the ideal gas law
PV =nRT
P =( 1 x 0.0821 x 292.15) ÷ 0.5
= 4.797 atm
the difference between the ideal pressure and real pressure is
Pressure by vander waal equation- Pressure by ideal gas law
45.88 - 4.797
= 41.083 atm.is the difference between the two.
The molecular formula for aspartame is C14H18N2O5, and its molar mass is about 294 g/mol.
Convert 1.2 g into moles, which gives
1.2 g / 294 g/mol = 4.08 X 10-3 moles aspartame.
Since each mole of aspartame has 2 moles of nitrogen, you have 8.16 X 10-3 moles of N in your 1.2 grams of aspartame.
Finally, multiply that by Avogadro's number to get the number of N atoms:
8.16 X 10^-3 mol X 6.02 X 10^23 = 4.9 X 10^21 nitrogen atoms.
The answer for your question is <span>No. This is because in given conditions, it is not the most stable form of oxygen's element. It will not equate into zero because there will be charge remained after balancing the equation.
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