Answer:
if you are talking about miles the answer would be is 0.0001509932(divided by 160934)
if you are talking about mililiters it would be 24.3(multiply times one)
hope this helps
Answer:
0.009
Explanation:
Molarity, a measure of molar concentration of a substance is calculated thus;
Molarity = number of moles ÷ volume
According to the provided information, mass of KCl = 0.47g, volume of water = 700ml
Using mole = mass/molar mass
Molar Mass of KCl = 39.10 + 35.453
= 74.553g/mol
Mole = 0.47/74.55
Mole = 0.0063mol
Volume of water = 700ml = 700/1000 = 0.7L
Molarity = 0.006/0.7
Molarity = 0.00857
The value of molarity rounded to three decimal places (3 d.p) = 0.009
Answer:
(CH3)3C^+ + OH^- --------> (CH3)3COH
Explanation:
This reaction has to do with SN1 reaction of alkyl halides. Here tert-butanol is formed from tert-butyl bromide.
The first step in the reaction is the formation of a carbocation. This is a unimolecular reaction. The rate of reaction depends on the concentration of the alkyl halide. This is a slow step and thus the rate determining step in the mechanism.
(CH3)3CBr -------> (CH3)3C^+ + Br^-
The second step is a fast step and it completes the reaction mechanism. It is a bimolecular reaction as follows;
(CH3)3C^+ + OH^- --------> (CH3)3COH
The balanced equation for combustion is as follows;
2CH₃OH + 3O₂ ---> 2CO₂ + 4H₂O
The stoichiometry of CH₃OH to O₂ is 2:3
the limiting reagent is the reactant that is fully consumed during the reaction. The amount of product formed is directly proportional to the amount of limiting reactant produced. The excess reagent is the reactant that is provided in excess and is not fully used up, there will be an amount of this reagent remaining after the reaction.
If methanol is the limiting reactant,
If 2 mol of methanol reacts with 3 moles of O₂
Then 24 mol of methanol reacts with - 3/2 x 24 = 36 mol of O₂ should be present
But only 15 mol of O₂ is present, therefore O₂ is the limiting reactant and methanol is in excess.
3 mol of O₂ reacts with 2 mol of CH₃OH
then 15 mol of O₂ reacts with 2/3 x 15 = 10 mol of CH₃OH
Excess reactant is methanol, 10 mol are used up therefore 24 - 10 mol = 14 mol are remaining at the end of the reaction
Answer:
Every 30 days hope this helps
