Explanation:
The given data is:
The half-life of gentamicin is 1.5 hrs.
The reaction follows first-order kinetics.
The initial concentration of the reactants is 8.4 x 10-5 M.
The concentration of reactant after 8 hrs can be calculated as shown below:
The formula of the half-life of the first-order reaction is:
Where k = rate constant
t1/2=half-life
So, the rate constant k value is:
The expression for the rate constant is :
Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:
Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.
Answer: -
The first step involves protonation of the carbonyl oxygen.
After protonation, the Alcohol oxygen now attacks the carbon of the carbonyl.
Thus a six membered ring is formed with 5 carbon atoms and 1 oxygen atom. The 1st position carbon atom has 2 OH groups.
One of these gets again protonated.
This leaves as water. With the loss of the H+, there results a carbonyl at 1 position.
Thus 5-hydroxypentanoic acid forms a lactone or 2-oxanone in presence of acid.
Answer:
CuSO4
Explanation:
Na2S + CuSO4 → Na2SO4 + CuS
The reaction is balanced (same number of elements in each side)
To determine limiting reagent you need to know the moles you have of each.
Molar mass Na2S = 23 * 2 + 32 = 78
Molar mass CuSO4 = 63.5 + 32 + 16 * 4 = 159.5
Na2S mole = 15.5 / 78 = 0.2
CuSO4 mole = 12.1/159.5 = 0.076
*Remember mole = mass / MM
With that information now you have to divide each moles by its respective stoichiometric coefficient
Na2S stoichiometric coefficient : 1
Na2S : 0.2 / 1 = 0.2
CuSO4 stoichiometric coefficient: 1
CuSO4: 0.076 / 1 = 0.076
The smaller number between them its the limiting reagent, CuSO4
Answer:
Quantity A is weight and Quantity B is mass
Explanation: weight has same unit as force. Mass is the quantity of matter present in a body or object
