Answer:
43.868 J
Explanation:
Kinetic energy of a body is the amount of energy possessed by a moving body. The SI unit of kinetic energy is the joule (kg⋅m²⋅s⁻²).
According to classical mechanics, kinetic energy = 1/2 m·v²
Where, m= mass in kg and v= velocity in m/s
Given: m = 19.2 lb and v = 7.10 miles/h
Since, 1 lb= 0.453592 kg
∴ m = 19.2 lb = 19.2 × 0.453592 kg = 8.709 kg
Also, 1 mi = 1609.34 m and 1 h = 3600 sec
∴ v = 7.10 mi/h = 7.10 × 1609.34 m ÷ 3600 sec = 3.174 m/sec
Therefore, <u>kinetic energy of the goose</u> = 1/2 m·v² = 1/2 × (8.709 kg)× (3.174 m/sec)² = 43.868 J
Answer:
![\frac{[F^{-}]}{[HF]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BF%5E%7B-%7D%5D%7D%7B%5BHF%5D%7D)
is larger
Explanation:
, where 
is the acid dissociation constant.
For a monoprotic acid e.g. HA, ![K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the value , lower will the the 
In this mixture, at equilibrium, ![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
will be constant.
of HF is grater than of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So, is larger
that's your work bro do by your own
Answer:
The molar mass of the gas is 44 g/mol
Explanation:
It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:
If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):
6,61 = √M₂
44g/mol = M₂
<em>The molar mass of the gas is 44 g/mol</em>
<em></em>
I hope it helps!
Answer:
In hyphen notation, the mass number is written after the name of the element. For example, in isotopic notation, the isotope of carbon that has a mass number of twelve would be represented as 12C . In hyphen notation, it would be written as carbon-12.
Explanation:
lol just took the question and looked it up online this was the first thing that i saw if its not correct im sorry
