Answer:
The reaction rate of the both questions remain unchanged.
Explanation:
For question 1: The reaction 1-iodo -2- methylbutane with cyanide ion is an SN2 reaction because the Alkyl halide is a primary alkyl halide. The rate of reaction is dependent on concentration of the nucleophile and the alkyl halide at the same. For the rate of reaction to be affected (increased or decreased), the concentration of nucleophile and the alkyl halide have to be altered.
For question 2: The reaction of 2-iodo -2- methylbutane with ethanol is an SN1 reaction because the Alkyl halide is a tertiary alkyl halide. There are two-step reaction mechanism in this reaction. The first step is the rate determining step which determines the extent of the reaction and hence the rate of reaction. For the rate of reaction to be affected (increased or decreased), the concentration of the Alkyl halide alone will be altered. The rate of reaction is independent of the concentration of the nucleophile.
Hello, 3Coli Here!
Your Answer is Here:
I would recheck, redo, and revise my answers, if necessary. Also, I can look at the lesson again, if I forgot something. So that's what I would do.
Hopefully, this helps!
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Answer:
MnSO₄.7H₂O
Explanation:
To solve this question, we need to convert the mass of the dehydrated MnSO₄. The difference between mass of the hydrate and dehydrated compound is the mass of water. With the mass we can find the moles of water and the formula of the hydrate:
<em>Moles MnSO₄ -Molar mass: 151g/mol-:</em>
17.51g * (1mol / 151g) = 0.116 moles
<em>Moles H₂O -Molar mass: 18g/mol-:</em>
32.14g-17.51g = 14.63g * (1mol / 18g) = 0.813 moles
The ratio of moles MnSO₄: Moles H₂O represent the amount of water molecules in the hydrate:
0.813mol / 0.116mol = 7 molecules of water.
The hydrate formula is:
<h3>MnSO₄.7H₂O</h3>
The answer is d. particles move from an area of high concentration to low concentration.
To solve this problem, we use Beer's Law: A= ε.l.c
A is the absorbance- 0,558
<span>ε is</span> the molar absorptivity- is <span>15000 </span><span><span>L⋅mol-1</span><span>cm-1</span></span>
<span>l is </span>the length of the cuvette- 1 cm
<span>c is</span> the molar concentration
Applying the formula,
0,558= 15000 x 1 x c
0,558/15000= c
c= <span>3.72×<span>10⁻⁵ </span> <span>mol⋅L<span>⁻¹</span></span></span>
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