Potential energy is energy due to an object's height above the ground.
Potential energy = mass x gravity x height
Kinetic energy is energy due to the motion of the object.
Kinetic energy = 1/2 x mass x velocity²
1.
The ball is not moving and is at a height above the ground so it has only potential energy.
P.E = 2 x 9.81 x 40
P.E = 784.8 J
2.
The ball is moving and has a height above the Earth's surface so it has both kinetic and potential energy.
P.E = same as part 1 = 784.8 J
K.E = 1/2 x 2 x 5²
K.E = 25 J
3.
The ball has no height above the Earth's surface and is moving so it has only kinetic energy.
K.E = 1/2 x 2 x 10²
K.E = 100 J
4.
50000 = 1/2 x 1000 x v²
v = 10 m/s
5.
39200 = 200 x 9.81 x h
h = 20.0 m
6.
12.5 = 1/2 x 1 x v²
v = 5 m/s
98 = 1 x 9.81 x h
h = 10.0 m
The volume of HCl required is 23.89 mL
Calculation of volume:
The reaction:

As HCl and NaOH react in 1 : 1 ratio.
Volume of NaOH= 718 mL
Concentration= 0.183M
Volume of HCl= ?
Concentration= 5.50M
Using the dilution formula:


Therefore,
Volume of HCl required will be 23.89 mL.
Learn more about neutralization reaction here,
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Answer:
(C) through the atmosphere
Explanation:
Answer:
The molality of isoborneol in camphor is 0.53 mol/kg.
Explanation:
Melting point of pure camphor= T =179°C
Melting point of sample =
= 165°C
Depression in freezing point = 

Depression in freezing point is also given by formula:

= The freezing point depression constant
m = molality of the sample
i = van't Hoff factor
We have:
= 40°C kg/mol
i = 1 ( organic compounds)



The molality of isoborneol in camphor is 0.53 mol/kg.
Answer:
The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²
Explanation:
The given parameters is as follows;
The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g
The required information = The number of formula units of potassium chloride (KCl)
The Molar Mass of KCl = 74.5513 g/mol

Therefore, we have;

1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of formula units
Therefore;
0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units
From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.