Answer:
The kinetic energy of an object is also measured in joules. Anything that is moving has kinetic energy, but various factors affect how much kinetic energy an object has. The first factor is speed. If two identical objects are moving at different speeds, the faster object has more kinetic energy. In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.
Explanation:
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)
Answer:
20 mL OF 6 M HYDROCHLORIC ACID WILL BE NEEDED
Explanation:
M1 V1 = M2 V2
M1 = Molarity of sodium hydroxide = 3 M
V1 = volume of sodium hydroxide = 40 mL
M2 = Molarity of hydrochloric acid = 6 M
V2 = Volume of hydrochloric acid = unknown
Rearranging the equation, we have:
V2 = M1 V1 / M2
V2 = 3 * 40 mL / 6
V2 = 120 / 6
V2 = 20 mL
To precipitate the benzoic acid by 6 M of hydrochloric acid, 20 mL volume will be needed.
