Answer:
Temperature decreases and density increases
Explanation:
Let us remember that density of a material increases as the temperature of the material decreases. So the cooler a material becomes, the denser it becomes also.
Between points B and C, the material rapidly cools down and the temperature decreases accordingly. This ultimately results in an increase in density since cooler materials are denser than hot materials.
Answer:
B.
General Formulas and Concepts:
<u>Chemistry - Organic</u>
- Hydrocarbons
- Bond-Line Notation: Each "arrow" point represents a carbon and the lines represent a bond between the carbons. Hydrogens are assumed added when drawing the molecular formulas
Explanation:
We are given C₈H₁₈. We need a Bond-Line notation where there are 8 "arrow" points (which represent Carbon). We can disregard the Hydrogens as they are assumed to take the rest of the bonds.
Only option B has a value of 8 carbons, represented by the "arrow" points. Therefore, it is the correct answer.
Answer:
Part a: The rate of the equation for 1st order reaction is given as ![Rate=k[H_2O_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BH_2O_2%5D)
Part b: The integrated Rate Law is given as ![[H_2O_2]=[H_2O_2]_0 e^{-kt}](https://tex.z-dn.net/?f=%5BH_2O_2%5D%3D%5BH_2O_2%5D_0%20e%5E%7B-kt%7D)
Part c: The value of rate constant is 
Part d: Concentration after 4000 s is 0.043 M.
Explanation:
By plotting the relation between the natural log of concentration of
, the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.
Part a
Rate Law
The rate of the equation for 1st order reaction is given as
![Rate=k[H_2O_2]](https://tex.z-dn.net/?f=Rate%3Dk%5BH_2O_2%5D)
Part b
Integrated Rate Law
The integrated Rate Law is given as
![[H_2O_2]=[H_2O_2]_0 e^{-kt}](https://tex.z-dn.net/?f=%5BH_2O_2%5D%3D%5BH_2O_2%5D_0%20e%5E%7B-kt%7D)
Part c
Value of the Rate Constant
Value of the rate constant is given by using the relation between 1st two observations i.e.
t1=0, M1=1.00
t2=120 s , M2=0.91
So k is calculated as

The value of rate constant is 
Part d
Concentration after 4000 s is given as

Concentration after 4000 s is 0.043 M.
To know the exact amount of sodium hydroxide that the teacher needs to order, we need to know how many students are there and the amount that each student uses.
We will then multiply these two values and get the amount needed to be ordered.
Since you have not provided such data in your question, I will just assume them to show the steps of the solution. You can then apply these steps to the values you have.
Now, assume that a class of 60 students and that each student needs to use 130 grams of sodium hydroxide to use during the lab.
This means that the teacher will order:
130 x 60 = 7800 grams = 7.8 kg of sodium hydroxide
B) <span>It is the instantaneous speed of the bike rider.</span>