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3 years ago

Who performed the oil-drop experiment?

Chemistry

2 answers:

evablogger [386]3 years ago

7 0

In the year 1909, Robert A. Millikan and Harvey Fletcher performed the oil drop experiment.

faltersainse [42]3 years ago

7 0

Answer:

Robert Millikan

Explanation:

(APEX)

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Name the three states of matter, and the<br> physical changes that occur between them.

Nataly [62]

Answer:

Solid, Liquid, Gas

Solid to Liquid - Melting

Solid to Gas - Sublimation

Liquid to Solid - Freezing

Liquid to Gas - Vaporization

Gas to Solid - Deposition

Gas to Liquid - Condensation

Explanation:

8 0

3 years ago

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Which body part MOST LIKELY formed this fossil?<br> blood<br> B<br> bones<br> fur<br> D<br> skin

Inessa [10]

Answer:

Bones are the correct answer.

8 0

3 years ago

What is the density of an object that has a mass of 149. 8 g and displaces 12. 1 ml of water when placed in a graduated cylinder

kozerog [31]

Answer:

Explanation:

The density of a substance can be found by using the formula

$d = \frac{m}{v} \\$

m is the mass

v is the volume

From the question

m = 149.8 g

v = 12.1 mL

We have

$d = \frac{149.8}{12.1} = 12.380165... \\$

We have the final answer as

Hope this helps you

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2 years ago

What makes the periodic table such a useful tool? *

Sidana [21]

Answer:

I think it is D. It allows you to compare the properties of the elements. I might be wrong.

5 0

3 years ago

A 1.93-mol sample of xenon gas is maintained in a 0.805-L container at 306 K. Calculate the pressure of the gas using both the i

Alex

Answer : The pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

Explanation :

First we have to calculate the pressure of gas by using ideal gas equation.

$PV=nRT$

where,

P = Pressure of $Xe$ gas = ?

V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

Now put all the given values in above equation, we get:

$P\times 0.805L=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=60.2atm$

Now we have to calculate the pressure of gas by using van der Waals equation.

$(P+\frac{an^2}{V^2})(V-nb)=nRT$

P = Pressure of $Xe$ gas = ?

V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

a = pressure constant = $4.19L^2atm/mol^2$

b = volume constant = $5.11\times 10^{-2}L/mol$

Now put all the given values in above equation, we get:

$(P+\frac{(4.19L^2atm/mol^2)\times (1.93mole)^2}{(0.805L)^2})[0.805L-(1.93mole)\times (5.11\times 10^{-2}L/mol)]=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=44.6atm$

Therefore, the pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

5 0

3 years ago