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2 years ago

2. A 10 kg ball is rolled down a rail from a height of 20 m. What is the velocity at the bottom of the

Chemistry

1 answer:

solmaris [256]2 years ago

3 0

Answer:

19.8m/s

Explanation:

Given parameters:

Mass of the ball = 10kg

Height of the rail = 20m

Unknown:

Velocity at the bottom of the rail = ?

Solution:

The velocity at the bottom of the rail is its final velocity.

Using the appropriate motion equation, we can find this parameter;

V² = U² + 2gH

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height

the ball was rolled from rest, U = 0

V² = O² + 2 x 9.8 x 20

V = 19.8m/s

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Since volume and temperature are constant, this means that pressure and number of moles are directly proportional. the sample with the largest number of moles will have the high pressure.

Since, the ideal gas equation is also called ideal gas law. So, according to ideal gas equations,

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At constant volume and temperature the equation become ,

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1 year ago

If you need to reverse the following reaction in order for it to be an intermediate reaction in a Hess's law problem, what would

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2 years ago

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19. What is the density of a 16.39

slavikrds [6]

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2 years ago

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2 years ago

The freezing point of a nonelectrolyte solution containing 30.0 g of a solute dissolved in 250.0 g of water is observed to be -2

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Answer:

Molar mass of solute is 89.28 g/m

Explanation:

Colligative property of freezing point depression to solve this:

ΔT = Kf . m . i

i = number of particles, dissolved in solution. In this case, it is a nonelectrolyte, so i = 1.

m = molalilty (mol of solute/1kg of solvent

ΔT = T° freeze pure solvent - T° freeze solution

0°C - (-2.50°C) = 1.86 °C/m . m

2.50°C / 1.86 m/°C = m

1.34 mol solute/kg solvent = m

This means, that in 1000 g of solvent, we have 1.34 moles but we have 250 g of solvent, so let's make a rule of three.

1000 g ____ 1.34 moles

250 g _____(2.50 . 1.34) / 1000 = 0.336 moles

To find the molar mass, we divide mass / moles

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2 years ago