stry: Chap x Mast gCh https:// gch stry.com/ Chemistry Section 14 Chapter 2 Home

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stry: Chap x Mast gCh https:// gch stry.com/ Chemistry Section 14 Chapter 2 Homework Exercise 2.80 Mercury is often used as an expansion medium in a thermometer. The mercury sits in a bulb on the bottom of the thermometer and rises up a thin capillary as the temperature rises. Suppose a mercury thermometer contains 3.350 g of mercury and has a capillary that is 0.220 mm in diameter. Lead metal can be extracted from a mineral called which contains 86.6% lead by mass. particular ore galena contains 68.5% galena by mass e QA search 54455010 Signed in as Sami Alzahrani I Help close Resources previous l 22 of 26 l next Part A How far does the mercury rise in the capillary when the temperature changes from 0.0 oC to 25.0 °C? The density of mercury at these temperatures is 13.596 g/cm3 and 13.534 g/cm3, respectively Express your answer with the appropriate units Ah 34857 cm. Submit My Answers Give U Incorrect, Try Again, 4 attempts remaining Part B h 92.5% If the lead can be extracted wi efficiency, what mass of ore is required to make a lead sphere with a 7.00 cm radius? Express your answer with the appropriate units 29.653 mole Subm My Answers Give U

Explanation / Answer

Part-A:

From given data,

mercury thermometer contains 3.250 g of mercury and has a capillary that is 0.200 mm in diameter.

The density of mercury at these temperatures is d1 = 13.596 gcm 3 , d2= 13.534 gcm 3.

T1 = 00C, T2 = 250C

V1 = m/d1 = 3.250 / 13.596
V2  =m/d2= 3.250 / 13.534
change in V = (3.250 / 13.596) - (3.250 / 13.534)

Volume of cylinder ( cap. tube) = pi r^2 h

From given data diameter = 0.200 mm

Hence radius = 0.200mm/ 2 = 0.100 mm = 0.0100 cm

(3.250 / 13.534) - (3.250 / 13.596) = pi (0.0100 ^ 2)h
If solved above equation for h, we will get

h = 3.4857 cm or 34.857 mm

Part-B:

Volume of the sphere = 4/3 * pi * 7.00^3 = 1436 cm^3

The density of lead is 11.34 grams / cm^3

The mass of the lead required to occupy the volume = 1436 x 11.34 = 16284.2 grams.

Now we have to find out what will produce that. Start with the Galena. It is 86.6 % lead.

86.6 % galena = 16284 grams
100% = x

x = 16284 /0.866 = 18804 grams of pure galena is required.

But the stuff you have is only 68.5% pure.

68.5 % = 18804 grams
100 % = x

x = 27451 grams of galena.

The process you are using is only 92.5 % efficient.

92.5 % = 27451 grams
100 % = x

x = 29676 grams of galena.
No. of moles of lead = 29676 g/207.2 = 143.2 moles.

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