Potential energy is stored energy. Kinetic energy involves movement.
If a ball is on the top of a hill, it has the most potential energy on the very top of the hill. The kinetic energy is also 0 at this point.
If the ball rolls down the hill, potential energy decreases while kinetic increases.
simple example
Answer:
27.4 gram is the solution it's simple dude...
Explanation:
don't be afraid of huge question they confuse you you need not to be confused
now see simple solution
molality is denoted by m
so
m= moles of solute / mass of solvent in kg.
i hope your know the meaning of solute and solvent....
so moles are given 0.467
and molar mass is given 58.44
so just take out the gram means
by applying formula
58.44×0.467
it will give 27.4 grams simple.....
Answer: option C. HF
Explanation: A polar bond is a covalent bond between two atoms where the electrons forming the bond are unequally distributed. Fluorine is more electronegative than hydrogen so the electrons in the bond are more closely associated with the fluorine atom than with the hydrogen atom.
Answer:
particles in 2 moles.
Explanation:
The number of particles that are contained in one mole, the international unit of amount of substance: by definition, exactly 6.022×10²³, and it is dimensionless. It is named after the scientist Amedeo Avogadro.
It is also known as Avogadro's constant.
∴ Number of particles in one mole = 
∴ Number of particles in 2 mole = 2 times Number of particles in one mole
∴ Number of particles in 2 mole=
Hence there are particles in 2 moles.
Answer:
Rate of formation of SO₃ ![[\frac{d[SO_{3}] }{dt}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%5D)
= 7.28 x 10⁻³ M/s
Explanation:
According to equation 2 SO₂(g) + O₂(g) → 2 SO₃(g)
Rate of disappearance of reactants = rate of appearance of products
⇒ ![-\frac{1}{2} \frac{d[SO_{2} ]}{dt} = -\frac{d[O_{2} ]}{dt}=\frac{1}{2} \frac{d[SO_{3} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B2%7D%20%5D%7D%7Bdt%7D%20%3D%20-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7Bd%5BSO_%7B3%7D%20%5D%7D%7Bdt%7D)
-----------------------------(1)
Given that the rate of disappearance of oxygen = ![-\frac{d[O_{2} ]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_%7B2%7D%20%5D%7D%7Bdt%7D)
= 3.64 x 10⁻³ M/s
So the rate of formation of SO₃ = ?
from equation (1) we can write
![\frac{d[SO_{3}] }{dt} = 2 [-\frac{d[O_{2}] }{dt} ]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D%20%3D%202%20%5B-%5Cfrac%7Bd%5BO_%7B2%7D%5D%20%7D%7Bdt%7D%20%5D)
⇒ ![\frac{d[SO_{3}] }{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BSO_%7B3%7D%5D%20%7D%7Bdt%7D)
= 2 x 3.64 x 10⁻³ M/s
⇒ = 7.28 x 10⁻³ M/s
∴ So the rate of formation of SO₃ = 7.28 x 10⁻³ M/s
