Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
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Answer:
his is an example of a first-year chemistry question where you must first convert two of the pressures to the units of the third and add them up, per Dalton’s law of additive pressures. There are three possible answers, one for each of the three pressure units.
1 atm = 760 torr …… torr and mm Hg are the same
1 atm = 101.3 kPa
Dalton’s law:
P(total) = P(O2) + P(N2) + P(CO2)
Explanation:
Gases will assume whatever pressure depending on the equation of state of the mixture (in this case) and the volume htey are contained in. That could be the ideal gas law and simple mixing law, If you are quoting the partial pressures which you call simply “the pressure” of each gas, and that these refer to their values in the present mixture, then yes, we would add them up. The pressures are low enough for the ideal gas law to apply provided the temperature is not extremely low as well .
Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:
Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:
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