PLA is a type of condensation polymer. Hydroxyl group of one lactic acid reacts with carboxylic group of second lactic acid and a ester is formed with the elimination of water molecule. And the reaction propagates with the increase in chain. And, the product formed is PLA as shown below.
Answer: d. Remove one-half of the initial CaCO3.
Explanation: Le Chatelier's principle states that changes on the temperature, pressure, concentration and volume of a system will affect the reaction in an observable way. So in the reaction above:
A decrease in temperature will shift the equilibrium to the left because the reaction is exothermic, which means heat is released during the reaction. In other words, when you decrease temperature of a system, the equilibrium is towards the exothermic reaction;
A change in volume or pressure, will result in a production of more or less moles of gas. A increase in volume or in the partial pressure of CO2, the side which produces more moles of gas will be favored. In the equilibrium above, the shift will be to the left.
A change in concentration will tip the equilibrium towards the change: in this system, removing the product will shift the equilibrium towards the production of more CaCO3 to return to the equilibrium.
So, the correct answer is D. Remove one-half of the initial CaCO3.
Calculate the normality of 1 Kg of aluminum sulfide in 5000 ml of solution.
Normality comes out to be 8.11
<h3>
Given </h3>
- Mass of solute: 1000g
- Volume of solution (V): 5000 ml = 5 liters
- Equivalent mass of solute (E) = molar mass / n-factor
n-factor for 
is 6 and molar mass is 148g
So, on calculating equivalent mass is equal to 24.66g
FORMULAE of Normality (N) = (Mass of the solute) / (Equivalent mass of the solute (E) × Volume of the solution (V)
N=
<u> N=8.11</u>
Therefore, normality of 1 kg aluminum sulfide is 8.11
Learn more about normality here brainly.com/question/25507216
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<h3><u>Answer</u>;</h3>
C3H4O
<h3><u>Explanation;</u></h3>
Empirical formula is the simplest formula of a compound;
Molar mass CO2 = 44.01
Mass of CO2 produced = 2.053 g
Mass of carbon in original sample = 12.01/44.01 × 2.053
= 0.5603g
Molar mass H2O = 18
Mass of H in original sample = 2/18 ×0.5601
= 0.0622 g
Thus; original sample contained 0.5603g C and 0.0622g H. The balance of the sample was O
Mass of O = 0.8715 - (0.5603 + 0.0622) = 0.249g
The mole ratio of C:H:O will be;
Moles C = 0.5603/12 = 0.0467
Moles H = 0.0622
Moles O = 0.249/16 = 0.01556
C:H:O = 0.0467:0.0622:0.01556
Divide through by 0.01556:
C:H:O = 3:4:1
Empirical formula is thus C3H4O
Well you know that you have to balance out the equation to 0. So if I is -1 and Ca is +2 you know that I had to equal to -2. So you can do that by having two of them. So Ca=+2 and I2=-2 so CaI2 would equal 0 meaning that would be the correct answer. Hope that I explained that correct :)
