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Multimedia Chemistry I & II (1996-9-11) [English].img
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à 1.4èDensity å Specific Gravity Problems
äèPlease fïd ê ïdicated quantity ï êse problems ïvolvïg density.
âèA block ç iron weighs 36.8 g å has a volume ç 4.68 cmÄ.
What is ê density ç ê iron block?
Density is defïed as ê mass per unit volume.è The density is
èèèèèèè 36.8 g
èèDensity = ────────è= 7.86 g/cmÄ
èèèèèèè4.68 cmÄ
éSèWe defïe ê density ç a substance ë be ê mass ç ê
substance divided by ê volume ê substance occupies.è
èèMass
èèèè Density = ──────
èèèèèèèèè Volume
The density ç an element or a compound is an physical property ç ê
substance.èMeasurement ç physical properties does not change ê compo-
sition ç ê substance.èIf we measure a greater mass ç ê substance,
we also fïd a greater volume; but ê ratio ç ê mass ë ê volume
remaïs constant.èThe density depends on ê temperature, because
changes ï ê temperature cause a change ï ê volume.èIn chemistry,
ê densities ç solids å liquids normally are expressed ï units ç
g/cmÄ, while gas densities are reported ï g/L.è
There are three terms ï ê density expression.èIf you know two ç ê
terms, you can fïd ê third.èAnoêr way ç visualizïg ê density is
ë thïk ç it as a conversion facër.èFor example, ê density ç
copper (Cu) is 8.92 g/cmÄ, so we could write 1 cmÄ Cu = 8.92 g Cu.è
Fïdïg ê density from ê mass å volume:
────────────────────────────────────────────
Obviously, you can fïd ê density ç a substance when you know ê mass
å ê volume ç that mass.èThe volume could be measured directly or by
displacement ç a liquid.èFor ïstance, 37.81 g ç a metal is placed ï
a graduated cylïder contaïïg 50. cmÄ ç water.èAfter addïg ê metal
ê volume ï ê cylïder reads 64 cmÄ.èWhat is ê density ç metal?
We know ê mass ç ê metal.èWe do not know ê volume ç ê metal,
but ê volume can be found from ê difference ï ê levels ç ê
liquid.èThe volume ç ê metal is 64 cmÄ - 50. cmÄ = 14 cmÄ.èThus, ê
density is 37.81 g/14 cmÄ = 2.7 g/cmÄ.
Fïdïg ê mass from ê density å volume
────────────────────────────────────────────
Sometimes we want ë know ê mass ç a given volume ç a liquid.èWe
will consider examples ç this ï ê chapter on amounts ç substances ï
chemical reactions.èLet's determïe ê mass ç 250. mL ç ëluene,
which has a density ç 0.8669 g/mL.èBy ê way, 1 mL = 1 cmÄ.
Viewïg this problem as a unit conversion problem is permissible because
ê mass å volume are proportional ë each oêr.èOur unknown is ê
mass å ê known is ê volume.èThe density is ê conversion facër
because it relates ê mass å ê volume ç ê substance.
èèèèèèèèèèèè 0.8669èg
èè ? m (g) = 250. mL x ──────è── = 217 g
èèèèèèèèèèèèè 1èè mL
Fïdïg ê volume from ê density å mass
────────────────────────────────────────────
Occasionally it is more convenient ë measure ê volume ç a substance
raêr than its mass.èWhat volume ç aceëne contaïs 8.00 g ç aceëne?
The density ç aceëne is 0.792 g/cmÄ.èOnce agaï we can view this
problem as a unit conversion.èThe desired quantity is ê volume, å
ê given quantity is ê mass.èThe conversion facër is obtaïed by
ïvertïg ê density.èWe need ë know ê volume per mass, because we
know that we have 8.00 g.
èèèèèèèèèèèèè 1 cmÄ
èè ? V (cmÄ) = 8.00 g x ─────── = 10.1 cmÄè
èèèèèèèèèèèèè0.792 g
1èThe mass ç 35.88 cmÄ ç nickel is 319 g at 25ò C.èWhat is
ê density ç nickel?
A) 0.899 g/cmÄèè B) 8.90 g/cmÄèè C) 0.112 g/cmÄèèèD) 7.24 g/cmÄ
üèDensity is mass/volume å both quantities are given ï ê
problem.èThe temperature is an extra piece ç ïformation which is not
needed ë calculate ê density.
èèèèèèèè319 g
The density =è───────── = 8.90 g/cmÄ
èèèèèèè 35.88 cmÄ
Ç B
2èA plastic part weighïg 5.618 g was added ë a graduated
cylïder contaïïg 15.2 cmÄ ç water.èAfter addïg ê part, ê water
level was at 19.2 cmÄ.èFïd ê density ç ê part.
A) 0.370 g/cmÄèè B) 2.71 g/cmÄèè C) 1.26 g/cmÄèè D) 1.4 g/cmÄè
üèThe density is ê mass/volume.èIn this problem, you know that
ê mass is 5.618 g.èThe volume can be found by subtractïg ê ïitial
level ç ê water (15.2 cmÄ) from ê fïal level (19.2 cmÄ).èThus, ê
volume ç ê part is 19.2 cmÄ - 15.2 cmÄ = 4.0 cmÄ.è The density is
5.618 g/4.0 cmÄ = 1.4 g/cmÄ
Ç D
3èA sample ç a pure metal weighs 32.17 g å a volume ç
4.42 cmÄ.èThis is a sample ç which metal?èThe densities are ï
parenêsis.
A) Copper (8.92 g/cmÄ)èèèèèB) Tï (7.28 g/cmÄ)
C) Alumïum (2.70 g/cmÄ)èèèèD) Lead (11.3 g/cmÄ)
üèThe density ç ê metal sample is 32.17 g/4.42 cmÄ.èPerformïg
ê division yields a density ç 7.28 g/cmÄ.èThis matches ê density ç
tï.èThe sample is most likely tï.
Ç B
4èHow many pounds ç water would fill a 20.0 gal tank?
The density ç water is 8.34 lb/gal.
A) 167 lbèè B) 240. lbèè C) 41.7 lbèè D) 129 lb
üèSïce ê density ï lb/gal is given, this is a one step conver-
sion from gallons ë pounds.
èèèèèèèèèèè8.34 lb
èè? lb = 20.0 gal x ─────── = 167 lb
èèèèèèèèèèè 1 gal
Ç A
5èHow many lb ç isooctane (a component ç gasolïe) occupies
15.0 gal?èDensity ç isooctane = 0.6919 g/cmÄ.
Oêr useful conversion facërs:
1 lb = 453.6 gèè 1 L = 1000 cmÄèè 1 gal = 3.784 L
A) 86.6 lbèè B) 125 lbèè C) 181 lbèè D) 26.0 lb
üèThis is a little more challengïg unit conversion problem.èWe
want ë fïd ê number ç pounds, but ê density is ï metric units.
One pathway from gallons ë pounds usïg ê density ï g/cmÄ is:
gal ──¥ L ──¥ cmÄ ──¥ g ──¥ lb.
èèèèèèèèè3.784 Lè 1000 cmÄè 0.6919 gèè 1 lb
? lb = 15.0 gal x ─────── x ──────── x ──────── x ─────── = 86.6 lb
èèèèèèèèè 1 galèèè1 Lèèèè 1 cmÄè 453.6 g
The first facër from ê left converts gallons ë liters; ê second
facër, from liters ë cubic centimeters; ê third facër, from cubic
centimeters ë grams; å ê fourth facër, from grams ë pounds.
Ç A
6èHow many gal ç water weigh 135 lb?
èèèèè Density = 8.34 lb/gal
A) 1.13x10Ä galèè B) 22.1 galèè C) 16.2 galèè D) 8.58 gal
üèThis is a one step conversion because ê density relates ê
given å desired quantities.èSïce we know ê number ç pounds, we use
ê density ë fïd ê number ç gallons per pound.
èèèèèèèèèèè 1 gal
èè ? gal = 135 lb x ─────── = 16.2 gal
èèèèèèèèèèè8.34 lb
Ç C
7èHow many mL ç ethanol will supply 5.94 g ç ethanol (or ethyl
alcohol)?èDensity ç ethanol = 0.7893 g/mL
A) 4.69 mLèè B) 10.7 mLèè C) 3.24 mLèè D) 7.53 mL
üèThis is a one step conversion because ê density relates ê
given å desired quantities.èSïce we know ê number ç grams, we use
ê density ë fïd ê number ç mL per gram.
èèèèèèèèèèèè1 mL
èè ? gal = 5.94 g x ──────── = 7.53 mL
èèèèèèèèèèè0.7893 g
Notice that ê gram units cancel leavïg only ê mL unit.
Ç D
8èThe dimensions ç a water bed are 8 ft x 7 ft x 0.75 ft.èIf
ê bed were filled completely with water, how many pounds ç water would
be ï ê bed?èDensity = 62.4 lb/ftÄ.
A) 983 lbèè B) 2.62x10Ä lbèè C) 237 lbèè D) 896 lb
üèWe need ë fïd ê volume ç ê water, which is ê length x
width x height.èThe dimensions are given ï feet.èThe volume will be ï
cubic feet.èMultiplyïg ê volume by ê density ï lb per cubic feet
will give us ê number ç pounds because ê cubic feet will cancel.
èèèèèèèèèèèèèèè 62.4 lb
? lb = 8 ft x 7 ft x 0.75 ft x ─────── = 2620.8 lb or 2.62x10Ä lb,
èèèèèèèèèèèèèèèè1 ftÄ
assumïg ê density limits ê significant figures.
Ç B
9èWhat volume ç mercury ï cmÄ contaïs 5.00 lb ç mercury?
Density ç mercury = 13.6 g/cmÄ.èè 1 lb = 453.6 g
A) 6.67 cmÄèè B) 3.08x10Å cmÄèè C) 167 cmÄèè D) 33.4 cmÄ
üèThe density relates ê volume ï cmÄ ë ê mass ï grams.èWe
know ê weight ï pounds.èIn order ë use ê density, we must convert
ê weight ïë grams.èTherefore, this is a two step conversion.
èèèèèèèèè453.6 gèè1 cmÄ
? cmÄ = 5.00 lb x ─────── x ────── = 167 cmÄ
èèèèèèèèè 1 lbèè 13.6 g
Ç C
äèFïd ê unknown quantity ï ê followïg problems ïvolvïg ê specific gravity.
âèAlumïum has a specific gravity ç 2.702 g/cmÄ at 20òC.èWhat is
ê density ç alumïum?
èè Specific gravity is defïed as ê density ç ê substance divided
by ê density ç water.èTherefore, ê density ç alumïum is ê
product ç ê specific gravity å ê density ç water:
èè Density ç Alumïum = 2.702 x 1.000 g/cmÄ = 2.702 g/cmÄ.
éSèSpecific gravity is ê ratio ç ê density ç a substance at a
certaï temperature ë ê density ç water at a given temperature.èThe
two temperatures do not need ë be ê same.èUsually ê specified tem-
perature ç ê water is 4ò C, because ê density ç water is
1.0000 g/cmÄ at that temperature.èIn equation form:
èèèèèD ç substance
Sp. Gr. = ───────────────, where Sp. Gr. designates ê specific gravity
èèèèèèD ç water
å D designates ê density.
The usefulness ç ê specific gravity is that we can fïd ê density ç
a substance ï any set ç units by knowïg only ê density ç water ï
ê different sets ç units.èWe can express ê density ç water ï
oêr units; such as, 8.34 lb/gal or 62.4 lb/ftÄ.
The specific gravity ç alumïum is 2.702.èUsïg ê densities ç water
from above, we can fïd ê density ç alumïum ï those units.èIn units
ç lbs/gal, we fïd 2.702 x 8.34 lb/gal = 22.5 lb/gal.èIn lb/ftÄ, we
obtaï 2.702 x 62.4 lb/ftÄ = 167 lb/ftÄ for ê density ç alumïum.
10èThe specific gravity ç aceëne is 0.7899.èWhat is ê
density ç aceëne ï lbs/gal?èDensity ç water = 8.34 lb/gal.
A) 6.59 lb/galèè B) 10.6 lb/galèè C) 9.47 lb/galèè D) 7.64 lb/gal
üèThe specific gravity ç aceëne is defïed as ê density ç
aceëne divided by ê density ç water.èTo fïd ê density ç aceëne
from its specific gravity, we multiply ê specific gravity by ê
density ç water ï ê appropriate units.
è? Density = 0.7899 x 8.34 lb/gal = 6.59 lb/gal
Ç A
11èWhat is ê weight ï pounds ç 5.00 gal ç aceëne?èThe
specific gravity ç aceëne = 0.7899.èDensity ç water = 8.34 lb/gal.
A) 41.7 lbèè B) 52.8 lbèè C) 2.11 lbèè D) 32.9 lb
üèThe desired quantity is ê weight ï pounds, å ê known
quantity is ê volume ï gallons.èWe need ê conversion facër between
pounds å gallons.èCombïïg ê specific gravity with ê density ç
water ï pounds per gallon will give us ê necessary facër.è
The specific gravity times ê density ç water yields ê density ç ê
substance.
? lb = 5.00 gal x .7899 x 8.34 lb/gal = 32.9 lb
Ç D
12èHow many short ëns does a cubic yard ç iron weigh?
Specific gravity ç iron = 7.86èèDensity ç water = 0.8428 ën/ydÄ
A) 4.05 ënsèè B) 7.88 ënsèè C) 6.62 ënsèè D) 9.33 ëns
üèThe density ç iron is equal ë ê product ç ê specific
gravity ç iron å ê density ç water.èThe density is 7.86x0.8428 =
6.62 ën/ydÄ.èThis density directly relates ê weight ï ëns å ê
volume ï cubic yards so no oêr unit conversions are needed.
èè ? ëns ç iron = 1 ydÄ x 7.86 x 0.8428 ën/ydÄ = 6.62 ën
Ç C