Answer:
1) No
2) The solvent contaminated the analyte
3) The solvent should be evaporated properly before washing and drying the analyte
Explanation:
During separation of organic compounds, solvents are used. These solvents are able to contaminate the analyte and lead to a large difference in melting point of solids obtained.
However, the error can be minimized by evaporating the solvent before washing, drying and melting point determination of the solid.
To figure out questions related to reacting moles/masses, the first step is always to write a complete balanced equation.
2Fe (s) + 3Cl2 (g) → 2FeCl3 (s)
Since Cl2 is the excess reactant, and Fe is the limiting reactant, we can simply find the number of moles of the product by comparing the mole ratio of the limiting reactant to the mole ratio of the product from the equation.
From the equation, mole ratio of Fe:FeCl3 = 2:2 = 1:1, the number of moles of product is exactly the same as the number of moles of the limiting reactant, which makes it 8 moles.
Note that if the mole ratio is not 1:1, you have to do some calculations to make sure the no. of moles is balanced at the end. Which means, if the mole ratio happened to be 1:2, the no. of moles of the product would be 8x2=16 instead.
So, your answer is 8 moles.
Answer:
3.82 x 10²¹ molecules As₂O₃
Explanation:
To find the amount of molecules arsenic (III) oxide (As₂O₃), you need to (1) convert kg to lbs, then (2) convert g As₂O₃ to moles As₂O₃ (via molar mass), and then (3) convert moles to molecules (via Avogadro's number).
1 kilogram = 2.2 lb
Molar Mass (As₂O₃): 2(74.992 g/mol) + 3(15.998 g/mol)
Molar Mass (As₂O₃): 197.978 g/mol
Avogadro's Number:
6.022 x 10²³ molecules = 1 mole
0.0146 g As₂O₃ 1 kg 189 lb
------------------------ x --------------- x ------------------ x ................
1 kg 2.2 lb
1 mole 6.022 x 10²³ molecules
x ------------------ x --------------------------------------- = 3.82 x 10²¹ molecules As₂O₃
197.978 g 1 mole
Answer:
The correct answer is option B.
Explanation:
As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.
After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.
