Answer:
44 g
Explanation:
The formula for the number of moles (n) is equal to 
.
Since we need to find the mass, we derive it from the formula of the number of moles and we get that mass = n x molecular weight .
The molecular weight of 
= 12 g/mol (from the carbon) + 19x4 g/mol (from the 4 fluorine atoms)= 88 g/mol
We plug in the numbers in the derived formula for the mass and we get :
mass = n x molecular weight = 0.5 mol x 88 g/mol = 44 g
Answer:
1.4 × 10^-4.
Explanation:
C3H6O3 + H2O <======> C3H5O3^- + H3O^+ ------------------------------------------(1).
So, from the question above we are given the following parameters or data which is going to help in solving this particular Question/problem;
=>concentration of the solution of lactic acid (CH3CH(OH)C00H) = 0.1 M and pH = 2.44.
Therefore, the concentration of the hydrogen ion[H^+} can be determined from the pH formula given below;
pH = - log { H^+}.
2.44 = - log { H^+}.
Therefore, {H^+} = 0.0036 M.
From the equation (1) given above, we have that the ratio for the equilibrium reaction is 1 : 1 : 1 :1. Therefore, molarity of C3H5O3^- = 0.0036 M and the molarity of C3H6O3 =( 0.1 - 0.0036 M) = 0.0964 M at equilibrium.
Hence, ka = {C3H5O3^-} { H3O^+} /{C3H6O3} = ( 0.0036 M)^2 /(0.0964 M) = 1.4 × 10^-4.
What's your question......................
Answer:
4) Titration
Explanation:
Titration is a common process used to determine the concentration of acids. It does this by adding a solution of base with a known concentration to the acid until it reaches neutralization.
Lets name the unknown metal as M. Cation would be M³⁺.
the molecular formula of the compound is M₂(SO₄)₃
the mass of one mole - (molar mass of M x2 + 3 x molar mass of SO₄²⁻)
= 2M + 96 x 3
= 2M + 288
In 1 mol if there's 72.07% of sulphate ,
then 72.07 % corresponds to 288 g
1 % is then - 288/72.07
100 % of the compound - 288/72.07 x 100
molar mass of the compound - 399.6 g/mol
mass of 2M - 399.6 - 288 = 111.6 g
molar mass of M - 111.6 /2 = 55.8 g/mol
the element with molar mass of 55.8 is Fe.
Unknown metal is iron(III) , Fe³⁺
