Fecl3 has a RFM of 56 + (35.5 * 3) = 162.5
so find out its moles 16.5 ÷ 162 5 = aprox. 0.1M
then there is a 1 : 3 ratio so 0.1 * 3 = 0.3
then times that by the avogado constant which is 6.02×10^23
0.3 × 6.02×10^23 = ans
hope that helps
Answer:
Esterification reaction
Explanation:
When we have to go from an acid to an ester we can use the <u>esterification reaction</u>. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).
In this case, we need the <u>methyl ester</u>, therefore we have to choose the <u>appropriate alcohol</u>, so we have to use the <u>methanol</u> as reactive if we have to produce the methyl ester.
ツ here your answer

- A)Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)
- 2KBr(aq)+BaI2(aq) → 2KI(aq)+BaBr2(s)
- B)Balance the Chemical Equation for the reaction of calcium carbonate with hydrochloric acid:
- CaCO3+ HCl -> CaCl2 + CO2 + H2O To balance chemical equations we need to look at each element individually on both sides of the equation. calcium carbonate is a chemical compound with the formula CaCO3.
<em><u>M</u></em><em><u>a</u></em><em><u>r</u></em><em><u>k</u></em><em><u> </u></em><em><u>m</u></em><em><u>e</u></em><em><u> </u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>b</u></em><em><u>r</u></em><em><u>a</u></em><em><u>i</u></em><em><u>n</u></em><em><u>l</u></em><em><u>i</u></em><em><u>s</u></em><em><u>t</u></em>
The correct answer from the choices given is the last option. The can from the <span> car will lose the carbon more quickly because there are fewer solute–solvent collisions. The can in the car has a lower temperature than the one in the refrigerator. At low temperature, the solubility of carbon dioxide in the liquid decrease therefore particles would tend to be in the vapor phase and escape from the liquid.</span>
The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:
0.115 mol I₂
1 - 0.115 = 0.885 mol CH₂Cl₂
We need moles of solute, which we have, and must convert our moles of solvent to kg:
0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂
We can now calculate the molality:
m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂
The molality of the iodine solution is 1.53.