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3 years ago

Stade avogadro's hypothesis what are its applications, prove that hydrogen hydrogen and oxygen gases

Physics

1 answer:

natita [175]3 years ago

3 0

Answer:

Applications of Avogadro's hypothesis:

In explaining Gay Lussac's law of gaseous volumes.

In determining the atomicity of gasses.

In determining the molecular formula of a gas.

In establishing the relationship between relative molecular mass and vapor density.

Explanation:

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A 4-kilogram ball moving at 8 m/sec to the right collides with a 1-kilogram ball at rest. After the collision,

elena-14-01-66 [18.8K]

Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s

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4 years ago

Help, please! Thank you for your kind gesture

blagie [28]

Answer:

The last option.

Explanation:

Since you are going down, the gravitational potential energy would go down too. Thus, the gravitational potential energy decreases.

Since the gravitational potential energy is converted to kinetic energy when you move down, there is an increase in kinetic energy.

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3 years ago

Read 2 more answers

which obstacle to problem solving involves arriving at a solution based on your preconceptions and ignoring any evidence that go

White raven [17]

Answer:

confirmation bias

Explanation:

the tendency to interpret new evidence as confirmation of one's existing beliefs or theories in a way that affirms one's prior beliefs or hypotheses.

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4 years ago

A mail carrier leaves the post ofﬁce and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00

Oxana [17]

Answer:

12.97 km

Explanation:

In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.

Taking north as positive y direction and east as positive x-direction, we have:

- Displacement 1: 2.00 km to the north

$A_x = 0\\A_y = +2.00 km$

- Displacement 2: 60.0° south of east for 7.00 km

$B_x=(7.00)(cos (-45^{\circ}))=4.95 km\\B_y = (7.00)(sin (-45^{\circ}))=-4.95 km$

- Displacement 3: 9.50 km 35.0° north of east

$C_x=(9.50)(cos 35^{\circ})=7.78 km\\C_y=(9.50)(sin 35^{\circ})=5.45 km$

So the net displacement along the two directions is:

$R_x=A_x+B_x+C_x=0+4.95+7.78=12.73 km\\R_y=A_y+B_y+C_y=2.00+(-4.95)+5.45=2.50 km$

So, the distance between the initial and final position is equal to the magnitude of the net displacement:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{12.73^2+2.50^2}=12.97 km$

7 0

3 years ago

In certain ranges of a piano keyboard, more than one string istuned to the same note to provide extra loudness. For example, the

steposvetlana [31]

Answer:

The beat frequency is 5.5 Hz.

Explanation:

f = 110 Hz, T = 600 N , T' = 540 N

let the frequency is f'.

$\frac{f'}{f}=\sqrt\frac{T'}{T}\\\\\frac{f'}{f}=\sqrt\frac{540}{600}\\\\\frac{f'}{f}=0.95\\\\f'= 104.5 Hz$

So, the beat frequency is f - f' = 110 - 104.5 = 5.5 Hz

6 0

3 years ago

Stade avogadro's hypothesis what are its applications, prove that hydrogen hydrogen and oxygen gases​

Stade avogadro's hypothesis what are its applications, prove that hydrogen hydrogen and oxygen gases