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3 years ago

What laws did dalton propose

Chemistry

1 answer:

Zina [86]3 years ago

6 0

Answer:

Dalton based his theory on two laws: the law of conservation of mass and the law of constant composition. The law of conservation of mass says that matter is not created or destroyed in a closed system.

Explanation:

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A 1.50-kilogram ball is attached to the end of a 0.520-meter string and swung in a circle. The velocity of the ball is 9.78 m/s.

Sedaia [141]

Answer:

F centripetal force (tension) = 275.9 N

Explanation:

Given data:

Mass = 1.50 kg

Radius = 0.520 m

Velocity of ball = 9.78 m/s

Tension = ?

Solution:

F centripetal force (tension) = m.v² / R

F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m

F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m

F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m

F centripetal force (tension) = 275.9 N

7 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:

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