39: An open-ended mercury manometer is used to measure the pressure exerted by a

Question

39: An open-ended mercury manometer is used to measure the pressure exerted by a trapped gas as shown in the figure. Atmospheric pressure is 750. mmHg. What is the pressure (in mmHg or torr) of the trapped gas if h =3 cm?

40: What volume (in mL) of 12 M HCI solution is required to prepare exactly 500. mL of a 0.38 M HCI solution?

50: When a 80 g sample of an alloy at 100.0 oC is dropped into 90.0 g of water at 22.1 oC, the final temperature is 34.0 oC. What is the specific heat of the alloy? (The specific heat of water is 4.184 J/(goC).)

53: When 193 J of energy are added to a sample of gallium that is initially at 25.0 oC, the temperature rises to 40.0 oC. What is the volume of the sample in cubic centimeter? Gallium, Ga specific hear 0.372 J(g^-1)(C^-1) Density 5.904 (g)(cm^-3)

Explanation / Answer

39)

given

height (h) = 3 cm

we know that

1 cm = 10 mm

so

h = 3 x 10

h = 30 mm

now

pressure by the gas = atomospheric pressure + height

pressure by the gas = 750 + 30

pressure by the gas = 780 mm

so

the pressure of the trapped gas is 780 mm

40)

we know that

for dilution

M1V1 = M2V2

given

initial concentration (M1) = 12

initial volumem (V1) = ???

final concentration (M2) = 0.38

final volume (V2) = 500 ml

so

12 x V1 = 0.38 x 500

V1 = 15.8333 ml

si

15.833 ml of 12 M HCl is required

50)

we know that

heat = mass x specific heat x temp change

so

heat gained by water = 90 x 4.184 x ( 34 - 22.1)

heat gained by water = 4481.064 J

now

we know that

heat lost by hot body = heat gained by cold body

in this case

heat lost by alloy = heat gained by water

so

heat lost by alloy = 4481.064 J

now

heat = mass x specific heat x temp change

so

80 x specific heat x ( 100-34) = 4481.064

specific heat of alloy = 0.8487

so

the specific heat of the alloy is 0.8487 J/g C

53)

now

heat = mass x specific heat x temp change

using given values

we get

193 = mass x 0.372 x ( 40-25)

193 = mass x 0.372 x 15

mass = 34.5878 g

now

we know that

volume = mass / density

given

density = 5.904

so

volume = 34.5878/5.904

volume = 5.8584

so

the volume of the sample is 5.8584 cm3

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