(a) Iron (iii) sulphate:
From the periodic table:
mass of iron = 55.845 grams
mass of sulphur = 32.065 grams
mass of oxygen = 16 grams
Iron (iii) sulphate has the formula: Fe2(SO4)3
molar mass = 2(55.845) + 3(32.065) + 3(4)(16) = 399.885 grams
(b) Sodium hydroxide:
From the periodic table:
mass of sodium = 22.989 grams
mass of oxygen = 16 grams
mass of hydrogen = 1 gram
Sodium hydroxide has the formula: NaOH
molar mass = 22.989 + 16 + 1 = 39.989 grams
(c) Barium carbonate
From the periodic table:
mass of barium = 137.327 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Barium carbonate has the formula: BaCO3
molar mass = 137.327 + 12 + 3(16) = 197.327 grams
(d) ammonium nitrate:
From the periodic table:
mass of nitrogen = 14 grams
mass of hydrogen = 1 gram
mass of oxygen = 16 grams
Ammonium nitrate has the formula: NH4NO3
molar mass = 14 + 4(1) + 14 + 3(16) = 80 grams
(e) Lead (iv) oxide
From the periodic table:
mass of lead = 207.2 grams
mass of oxygen = 16 grams
Lead (iv) oxide has the formula: PbO2
molar mass = 207.2 + 2(16) = 239.2 grams
From the above calculations, we can see that:
Iron (iii) sulphate has the greatest mass.
Acetic Acid has two C atoms, four H atoms, and two O atoms, so the molecular formula is C2H4O2.
Answer:
1.3×10⁻³ M
Explanation:
Hello,
In this case, given the dissociation reaction of acetic acid:

We can write the law of mass action for it:
![Ka=\frac{[H_3O^+][CH_3CO_2^-]}{[CH_3CO_2H]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BCH_3CO_2%5E-%5D%7D%7B%5BCH_3CO_2H%5D%7D)
Of course, excluding the water as heterogeneous substances are not included. Then, in terms of the change
due to the dissociation extent, we are able to rewrite it as shown below:

Thus, via the quadratic equation or solve, we obtain the following solutions:

Obviously, the solution is 0.00133M which match with the hydronium concentration, thus, answer is: 1.3×10⁻³ M in scientific notation.
Regards.
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K = 
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M