Answer:
single displacement reaction
Answer:
The empirical formula for the compound is C3H4O3
Explanation:
The following data were obtained from the question:
Carbon (C) = 40.92%
Hydrogen (H) = 4.58%
Oxygen (O) = 54.50%
The empirical formula for the compound can be obtained as follow:
C = 40.92%
H = 4.58%
O = 54.50%
Divide by their molar mass
C = 40.92/12 = 3.41
H = 4.58/1 = 4.58
O = 54.50/16 = 3.41
Divide by the smallest i.e 3.41
C = 3.41/3.41 = 1
H = 4.58/3.41 = 1.3
O = 3.41/3.41 = 1
Multiply through by 3 to express in whole number
C = 1 x 3 = 3
H = 1.3 x 3 = 4
O = 1 x 3 = 3
The empirical formula for the compound is C3H4O3
Answer:
Keq = [CO₂]/[O₂]
Explanation:
Step 1: Write the balanced equation for the reaction at equilibrium
C(s) + O₂(g) ⇄ CO₂(g)
Step 2: Write the expression for the equilibrium constant (Keq)
The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:
Keq = [CO₂]/[O₂]
Answer:
32g
Explanation:
We have to remember that for percent (w/w) concentration we usually write;
Percent concentration= mass of solute/mass of solution ×100
Since mass of solute= 14.7 g and percent concentration = 32.2%
Then
Mass of solution= mass of solute × 100/ percent concentration
Mass of solution= 14.7 ×100/32.2
Mass of solution= 46.7 g
Since mass of solution = mass of solute + mass of solvent
Mass of solute= 14.7 g
Mass of solution = 46.7g
Mass of solvent = 46.7g -14.7g = 32g
