The area is 60.2, to find area it's length times width.
Answer: 1.2642*10²⁵ on both sides
Explanation:
First check how many moles are there on each side.
Since this is a balanaced equataion the number of moles on each side is the same thus the number of atoms is also same on both sides
There are 3 moles of carbon and 8 moles of hydrogen in C3H8
and 2 moles of oxygen in O2 but there 5 infront so 2*5 is 10
Number of moles on the right is 10+8+3 = 21
Now use Avogrado's constant
21 Moles* (6.02*10²³)/Mol
= 21*6.02*10²³
= 1.2642*10²⁵
Answer:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
As molecules heat they begin to move faster. The heat from the metal plate will make the molecules at room temperature move faster. And the room temperature makes the hot place cool, making the hot plate molecules slow down.
To visualize this, you can use this link
https://phet.colorado.edu/en/simulations/gas-properties
click the play button to activate the activity
Answer:
Extensive properties vary with the amount of the substance and include mass, weight, and volume. Intensive properties, in contrast, do not depend on the amount of the substance; they include color, melting point, boiling point, electrical conductivity, and physical state at a given temperature.
Answer:
PH= 6.767 (answer is the A option)
Explanation:
first we need to correct the value in Kw at this temperature is 2.92*10^-14
so, in this case we have that:
Kw=2.92*10^-14 M²
[ H3O^+] [ H3O^+]
![[H_{3}O^{+} ] [OH^{-} ] = Kw = 2.92*10^{-14} M^{2} \\\\](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%5BOH%5E%7B-%7D%20%20%5D%20%3D%20Kw%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D%20%20%20%5C%5C%5C%5C)
at 40ºC
![[H_{3}O^{+} ] = [OH^{-} ]](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%5BOH%5E%7B-%7D%20%20%5D)
![[H_{3}O^{+} ]^{2} = 2.92*10^{-14} M^{2}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%5E%7B2%7D%20%3D%202.92%2A10%5E%7B-14%7D%20M%5E%7B2%7D)
![[H_{3}O^{+} ] = (2.92*10^{-14})^{1/2} = 1.71*10^{-7} M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20%282.92%2A10%5E%7B-14%7D%29%5E%7B1%2F2%7D%20%3D%201.71%2A10%5E%7B-7%7D%20M)
![PH= -log10[H_{3}O^{+} ] = -log10(1.71*10^{-7} ) = 6.767](https://tex.z-dn.net/?f=PH%3D%20-log10%5BH_%7B3%7DO%5E%7B%2B%7D%20%20%5D%20%3D%20-log10%281.71%2A10%5E%7B-7%7D%20%29%20%3D%206.767)