You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b
1 answer:
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Answer:
a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) ⇒ 16
c) No
was not the limiting reactant.
Explanation:
Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.
a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.
b) ⇒ 16
c) was not the limiting reactant based on the mol to mol ratio of
and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of
would also be produced.
Answer: 1. moles
2. 90 mg
Explanation:
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus moles of ozone is removed by =
moles of sodium iodide.
Thus moles of sodium iodide are needed to remove
moles of
2.
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by = moles of sodium iodide.
Mass of sodium iodide= (1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of .