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Concept explainers
(a)
Interpretation:
The quantity 4.5 cm has to be converted into oA.
(a)
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Answer to Problem 20PE
The quantity 4.5 cm in oA is 4.5×108 oA.
Explanation of Solution
The given quantity is 4.5 cm.
Centimeters is converted into oA using the conversion factor,
(1 m100 cm)(1 oA10-10 m)
The quantity 4.5 cm is converted as,
oA=4.5 cm(1 m100 cm)(1 oA10-10 m)oA=4.5×108oA
The quantity 4.5 cm in oA is 4.5×108 oA.
(b)
Interpretation:
The quantity 12 nm has to be converted into centimeters.
(b)
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Answer to Problem 20PE
The quantity 12 nm in centimeters is 1.2×10-6 cm.
Explanation of Solution
The given quantity is 12 nm.
Nanometers is converted into centimeters using the conversion factor,
(10-9 m1 nm)(100 cm1 m)
The quantity 12 nm is converted as,
cm=12 nm(10-9 m1 nm)(100 cm1 m)cm=1.2×10-6 cm
The quantity 12 nm in centimeters is 1.2×10-6 cm.
(c)
Interpretation:
The quantity 8.0 km has to be converted into millimeters.
(c)
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Answer to Problem 20PE
The quantity 8.0 km in millimeters is 8.0×106 mm.
Explanation of Solution
The given quantity is 8.0 km.
Kilometers is converted into millimeters using the conversion factor,
(1000 m1km)(1000 mm1 m)
The quantity 8.0 km is converted as,
mm=8.0km(1000 m1km)(1000 mm1 m)mm=8.0×106 mm
The quantity 8.0 km in millimeters is 8.0×106 mm.
(d)
Interpretation:
The quantity 164 mg has to be converted into grams.
(d)
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Answer to Problem 20PE
The quantity 164 mg in grams is 0.164 g.
Explanation of Solution
The given quantity is 164 mg.
Milligrams is converted into grams using the conversion factor,
(1 g1000 mg)
The quantity 164 mg is converted as,
g=164 mg(1 g1000 mg)g=0.164 g
The quantity 164 mg in grams is 0.164 g.
(e)
Interpretation:
The quantity 0.65 kg has to be converted into milligrams.
(e)
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Answer to Problem 20PE
The quantity 0.65 kg in milligrams is 6.5×105 mg.
Explanation of Solution
The given quantity is 0.65 kg.
Kilograms is converted into milligrams using the conversion factor,
(1000 g1kg)(1000 mg1 g)
The quantity 0.65 kg is converted as,
mg=0.65 kg(1000 g1kg)(1000 mg1 g)mg=6.5×105 mg
The quantity 0.65 kg in milligrams is 6.5×105 mg.
(f)
Interpretation:
The quantity 5.5 kg has to be converted into grams.
(f)
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Answer to Problem 20PE
The quantity 5.5 kg in grams is 5.5×103 g.
Explanation of Solution
The given quantity is 5.5 kg.
Kilograms is converted into grams using the conversion factor,
(1000 g1 kg)
The quantity 5.5 kg is converted as,
g=5.5 kg(1000 g1 kg)g=5.5×103 g
The quantity 5.5 kg in grams is 5.5×103 g.
(g)
Interpretation:
The quantity 0.468 L has to be converted into milliliters.
(g)
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Answer to Problem 20PE
The quantity 0.468 L in milliliters is 468 mL.
Explanation of Solution
The given quantity is 0.468 L.
Liters is converted into milliliters using the conversion factor,
(1000 mL1L)
The quantity 0.468 L is converted as,
mL=0.468 L(1000 mL1L)mL=468 mL
The quantity 0.468 L in milliliters is 468 mL.
(h)
Interpretation:
The quantity 9.0 μL has to be converted into milliliters.
(h)
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Answer to Problem 20PE
The quantity 9.0 μL in milliliters is 9.0×10-3mL.
Explanation of Solution
The given quantity is 9.0 μL.
Microliters is converted into milliliters using the conversion factor,
(1L106μL)(1000 mL1 L)
The quantity 9.0 μL is converted as,
mL=9.0 μL(1L106μL)(1000 mL1 L)mL=9.0×10-3 mL
The quantity 9.0 μL in milliliters is 9.0×10-3mL.
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Chapter 2 Solutions
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
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