For this item, we need to assume that air behaves like that of an ideal gas. Ideal gases follow the ideal gas law which can be written as follow,
PV = nRT
where P is the pressure,
V is the volume,
n is the number of mols,
R is the universal gas constant, and
T is temperature
In this item, we are to determine first the number of moles, n. We derive the equation,
n = PV /RT
Substitute the given values,
n = (1 atm)(5 x 10³ L) / (0.0821 L.atm/mol.K)(0 + 273.15)
n = 223.08 mols
From the given molar mass, we calculate for the mass of air.
m = (223.08 mols)(28.98 g/mol) = 6464.9 g
<em>ANSWER: 6464.9 g</em>
Answer:
Covalent bonding is the sharing of electrons between atoms. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small for ionic bonding to occur.
For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet.
Explanation:
Answer:
D is answer ..... !!!!!!!!.......!!!!!!#.#.....
Answer:
Explanation:
Hello,
In this case, considering the given reaction:
Thus, for the equilibrium, just water and hydrogen participate as iron and iron(II,III) oxide are solid:
![Kc=\frac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Thus, at the beginning, the concentration of water is 0.05 M and consequently, at equilibrium, considering the ICE procedure, we have:
Thus, the change 
is obtained as:
![\sqrt[4]{5.1} =\sqrt[4]{[\frac{(4x)}{(0.05M-4x)}]^4}\\\\1.5=\frac{(4x)}{(0.05M-4x)}\\\\x=0.0075M](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5.1%7D%20%3D%5Csqrt%5B4%5D%7B%5B%5Cfrac%7B%284x%29%7D%7B%280.05M-4x%29%7D%5D%5E4%7D%5C%5C%5C%5C1.5%3D%5Cfrac%7B%284x%29%7D%7B%280.05M-4x%29%7D%5C%5C%5C%5Cx%3D0.0075M)
Thus, the moles of hydrogen at equilibrium are:
![[H_2]_{eq}=4*0.0075\frac{mol}{L}*1.0L=0.03molH_2](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D4%2A0.0075%5Cfrac%7Bmol%7D%7BL%7D%2A1.0L%3D0.03molH_2)
Therefore, the grams of iron(II,III) oxide finally result:
Best regards.
Answer : The approximate atomic mass of an oxygen atom daltons (Da) is, 16 Da
Explanation :
As we know that dalton (Da) is equal to atomic mass unit (amu).
As we are given that,
Element = oxygen
Atomic number of oxygen = 8
As we know that the atomic mass of oxygen = 16 amu
So, the approximate atomic mass of an oxygen atom in daltons (Da) = Atomic mass unit of oxygen = 16 amu = 16 Da
Thus, the approximate atomic mass of an oxygen atom daltons (Da) is, 16 Da
