What is the name of the molecule shown below?
2 answers:
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Answer:
a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.
b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.
c)If concentration of is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d) If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.
Explanation:
Sucrose + fructose+ glucose
The rate law of the reaction is given as:
[sucrose]= 1.0 M
..[1]
a)
The rate of the reaction when [Sucrose] is changed to 2.5 M = R'
..[2]
[2] ÷ [1]
If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.
b)
The rate of the reaction when [Sucrose] is changed to 0.5 M = R'
..[2]
[2] ÷ [1]
If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.
c)
The rate of the reaction when is changed to 0.001 M = R'
..[2]
[2] ÷ [1]
If concentration of is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d)
The rate of the reaction when [sucrose] and both are changed to 0.1 M = R'
..[2]
[2] ÷ [1]
If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.
a. t=0.553 s
b. vox(horizontal speed) = 3.62 m/s
<h3>Further explanation</h3>Given
h = 1.5 m
x = 2 m
Required
a. time
b. vo=initial speed
Solution
Free fall motion
a. h = 1/2 gt²(vertical motion=h=voyt+1/2gt²⇒voy = 0)
t = √2h/g
t = √2.1.5/9.8
t=0.553 s
b. x=vox.t(horizontal motion)
vox=x/t
vox=2/0.553
vox=3.62 m/s