If the concentration of acetyl chloride is increased ten times the rate of reaction is increased ten times.
The conversion of acetyl chloride to methyl acetate is a substitution reaction. Recall that a substitution reaction is one in which a moiety in a molecule is replaced by another.
In this reaction, the CH3O- ion replaces the chloride ion. In the first step, the CH3O- ion attacks the substrate in a slow step. This creates a tetrahedral intermediate. Loss of the chloride ion yields the methyl acetate product.
The rate determining step is the formation of the tetrahedral intermediate. Since the reaction is first order in the acetyl chloride, if its concentration is increased ten times the rate of reaction is increased ten times.
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Answer:
D
Explanation:
31 / 2.8 = 11.0714286 L per mole of helium
3.5 / 4 = 0.875 moles
2.8 + 0.875 = 3.675 moles
11.0714286 x 3.675 = 40.6875 L
Answer:
Formula of magnesium oxide is MgO and formula of strontium oxide is SrO
Explanation:
Ca, Mg and Sr are same group bivalent metals.
Oxygen is a bivalent element
In general, formula of a compound made of a cation 
and an anion 
is 
for 
When x = y, formula of the compound becomes MA
As both these metals and oxygen are bivalent therefore formula of magnesium oxide is MgO and formula of strontium oxide is SrO.
To determine the amount of 6.0 M H2SO4 needed for the preparation, equate the number of moles of the 6.0 M and 2.5 M H2SO4 solution. This is done as follows
M1 x V1 = M2 x V2
Substituting the known variables,
(6.0 M) x V1 = (2.5 M) x (4.8 L)
Solving for V1 gives an answer of V1 = 2 L. Thus, to prepare the needed solution, dilute 2 L of 6.0 M H2SO4 solution with water until the volume reach 4.8 L.
M = n x Mr
Mr of H20 - 18.006
M = 1.15mol x 18.006g/mol
= 20.7069g
= 20.7g (3sfg)
