Hydrogen is composed of H atom and oxygen is composed of O atom. For water, it is composed by both H and O atom. If you burn hydrogen in oxygen, you can get water. And if you electrolysis water, you can get hydrogen and oxygen.
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: An increase in the ratio of insulin to glucagon will increase the activity of --
- Acetyl-CoA carboxylase(+)
-Phosphofructokinase PFK2(+)
-Glycogen synthase(+)
- Hormone sensitive lipase (-). The hormone sensitive lipase activity is not increased with increased insulin activity.
Explanation: increased insulin - glucagon ratio is usually high in fed state.Insulin helps the cells absorb glucose, reducing blood sugar and providing the cells with glucose for energy. When blood sugar levels are too low, the pancreas releases glucagon. Glucagon instructs the liver to release stored glucose, which causes blood sugar to rise.
Answer:
About 1.301 atm
Explanation:
The formula that you should is PV=nRT, where P stands for pressure, V stands for volume, n stands for the number of moles, R stands for the universal gas constant, and T stands for temperature in Kelvin. Since the volume, number of moles, and universal gas constant don't change, you don't need to worry about them.
1.07V=393nR
PV=498nR
P=1.301 atm. Hope this helps!
Answer:
Option 4
Explanation:
Hydrochloric acid is a strong one, that gives protons to medium. It can be dissociated as this:
HCl → H⁺ + Cl⁻
M means Molarity. It is a sort of concentration that indicates the moles of solute in 1L of solution.
M = moles / volume (L)
We can also say M = mmoles / mL of solution
M . mL = mmoles
0.453 M . 62.85mL = 28.5 mmoles
If we divide by 1000 → 28.5 mmol . 1 mol / 1000 mmol = 0.0285
