Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen 
if it occupies 31.6 mL at 
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.
Temperature = 
As molar mass of 
is 32 g/mol. Hence, the number of moles of are calculated as follows.
Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.
Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen if it occupies 31.6 mL at .
Answer:
atoms tend to react in order to gain 8 valence electrons
Explanation:
The octet rule describes the tendency of atoms of elements to react in order to have eight electrons in their valence shell. This is because having eight valence electrons confers stability to the atoms of these elements in the compounds they form.
The octet rule only does not apply to the transition elements or the inner transition elements as only the s and p electrons are involved. the electronic configuration in atoms having an octet is s²p⁶.
For example, sodium atom has one valence electron in its valence shell but a complete octet in the inner shell; it will react with chlorine atom which has seven valence electrons to form a stable compound, sodium chloride by donating its one valence electron in order to have an octet. Similarly, the chlorine atom will then have an octet by accepting the one electron from sodium atom.
Answer:
The density of the metal is 0.561 g/mL
Explanation:
The computation of the density of the metal is shown below;
As we know that
The Density of the metal is
where,
Mass = 4.9g
Change in volume = 6.9 mL
Now place these values to the above formula
So, the density of the metal is
= 0.561 g/mL
Hence, the density of the metal is 0.561 g/mL
We simply applied the above formula so that the correct density could arrive
Answer:- HBr is limiting reactant.
Solution:- The given balanced equation is:
From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.
The calculations could be shown as:
= 24 mol HBr
From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.
