Explanation:
Z = atomic mass of the element and , A = atomic mass of the element .
a) Z = 11, A = 23
Element = Sodium
symbol: ²³₁₁Na .
b) Z = 28, A = 64
Element = Nickel
symbol: ⁶⁴₂₈Ni .
c) Z = 50, A = 115
Element = tin
symbol: ¹¹⁵₅₀Sn .
d) Z = 20, A = 42
Element = Calcium
symbol: ⁴²₂₀Ca .
As we have the balanced reaction equation is:
N2O4 (g) ↔ 2NO2(g)
from this balanced equation, we can get the equilibrium constant expression
KC = [NO2]^2[N2O4]^1
from this expression, we can see that [NO2 ] is with 2 exponent of the stoichiometric and we can see that from the balanced equation as NO2
is 2NO2 in the balanced equation.
and [N2O4] is with 1 exponent of the stoichiometric and we can see that from the balanced equation as N2O4 is 1 N2O4 in the balanced equation.
∴ the correct exponent for N2O4 in the equilibrium constant expression is 1
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm
Ideal gas equation: pV = nRT => n = pV / RT
R = 0.0821 atm*liter/K*mol
V = 300 cm^3 = 0.300 liter
T = 298 K
p = 1 atm
=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol
2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type
X (+) + O2 (g) ---> X2O or
2 X(2+) + O2(g) ----> X2O2 = 2XO or
4X(3+) + 3O2(g) ---> 2X2O3
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)
So, lets probe those 3 cases.
3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol
=> x = 0.01226 moles of metal X
Now you can calculate the atomic mass of the hypotethical metal:
1.15 grams / 0.01226 mol = 93.8 g / mol
That does not correspond to any of the metal with valence 1+
So, now probe the case 2.
4) Case 2:
2moles X metal / 1 mol O2(g) = x / 0.01226 mol
=> x = 2 * 0.01226 = 0.02452 mol
And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol
That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.
4) Case 3
4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635
atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol
That does not correspond to any metal.
Conclusion: the identity of the metallic element could be titanium.
