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 ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
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:
![R=k[H^+][sucrose]](https://tex.z-dn.net/?f=R%3Dk%5BH%5E%2B%5D%5Bsucrose%5D)
![[H^+]=0.01M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01M)
[sucrose]= 1.0 M
![R=k[0.01M][1.0 M]](https://tex.z-dn.net/?f=R%3Dk%5B0.01M%5D%5B1.0%20M%5D)
..[1]
a)
The rate of the reaction when [Sucrose] is changed to 2.5 M = R'
![R'=[0.01 M][2.5 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.01%20M%5D%5B2.5%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.01%20M%5D%5B2.5%20M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
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'
![R'=[0.01 M][0.5 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.01%20M%5D%5B0.5%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.01%20M%5D%5B0.5%20M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
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'
![R'=[0.0001 M][1.0 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.0001%20M%5D%5B1.0%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.0001%20M%5D%5B1.0M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
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'
![R'=[0.1M][0.1M]](https://tex.z-dn.net/?f=R%27%3D%5B0.1M%5D%5B0.1M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.1M%5D%5B0.1M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.
Answer:
I can give you the definition ... That might help cause I honestly don't kno the answer either ;-;
Explanation:
When used as a diacritic mark, the term dot is usually reserved for the interpunct, or to the glyphs 'combining dot above' and 'combining dot below' which may be combined with some letters of the extended Latin alphabets in use in Central European languages and Vietnamese.
Here is an example:
The dot product between a unit vector and itself is also simple to compute. In this case, the angle is zero and cosθ=1. Given that the vectors are all of length one, the dot products are i⋅i=j⋅j=k⋅k=1.
Answer:
Theoretical yield for CO₂ is 5.10g
Explanation:
Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(g)
We convert the mass of oxygen to moles:
4.64 g /32 g/mol = 0.145 moles of O₂
Let's find out the 100% yield reaction of CO₂ (theoretical yield)
Ratio is 15:12. So let's make this rule of three:
15 moles of O₂ can produce 12 moles of CO₂
Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles
We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g
Answer:
created an organizational hierarchy
Explanation:
