The mass in kilogram has to be converted to micrograms. Concept Introduction: Conversion-factor method is the one which can be used to convert one metric unit into another. With this, any unit can be converted to another unit by means of ratio. The ratio that is used to convert unit is known as conversion-factor. For example, one kilogram can be converted into gram by multiplying with 1000.
The mass in kilogram has to be converted to micrograms. Concept Introduction: Conversion-factor method is the one which can be used to convert one metric unit into another. With this, any unit can be converted to another unit by means of ratio. The ratio that is used to convert unit is known as conversion-factor. For example, one kilogram can be converted into gram by multiplying with 1000.
The mass in kilogram has to be converted to micrograms.
Concept Introduction:
Conversion-factor method is the one which can be used to convert one metric unit into another. With this, any unit can be converted to another unit by means of ratio. The ratio that is used to convert unit is known as conversion-factor. For example, one kilogram can be converted into gram by multiplying with 1000.
(a)
Expert Solution
Answer to Problem 1.133QP
The given mass in kilogram is converted to micrograms as 8.45 ×109μg.
Explanation of Solution
Given mass in problem statement is 8.45 kg.
Since, 1kg=103g and 1μg=10-6g, we can use this to convert the above unit in micrograms,
8.45 kg × 103 g1 kg × 1 μg10-6 g=8.45×109μg
The given mass was converted into micrograms as shown above.
(b)
Interpretation Introduction
Interpretation:
The time in microseconds has to be converted to milliseconds.
Concept Introduction:
Conversion-factor method is the one which can be used to convert one metric unit into another. With this, any unit can be converted to another unit by means of ratio. The ratio that is used to convert unit is known as conversion-factor. For example, one kilogram can be converted into gram by multiplying with 1000.
(b)
Expert Solution
Answer to Problem 1.133QP
The given microseconds is converted to milliseconds as 3.18 ×10-1ms.
Explanation of Solution
Given time in problem statement is 318 μs.
Since, 1μs=10-6s and 1ms=10-3s, we can use this to convert the above unit in milliseconds,
318 μs × 10-6 s1 μs × 1 ms10-3 s=3.18×10-1ms
The given time was converted into milliseconds as shown above.
(c)
Interpretation Introduction
Interpretation:
The distance in kilometers has to be converted to nanometers.
Concept Introduction:
Conversion-factor method is the one which can be used to convert one metric unit into another. With this, any unit can be converted to another unit by means of ratio. The ratio that is used to convert unit is known as conversion-factor. For example, one kilogram can be converted into gram by multiplying with 1000.
(c)
Expert Solution
Answer to Problem 1.133QP
The given kilometers is converted to nanometers as 9.3 ×1013nm.
Explanation of Solution
Given distance in problem statement is 93 km.
Since, 1km=103m and 1nm=10-9m, we can use this to convert the above unit in nanometers,
93 km × 103 m1 km × 1 nm10-9 m=9.3×1013nm
The given distance was converted into nanometers as shown above.
(d)
Interpretation Introduction
Interpretation:
The distance in millimeters has to be converted to centimeters.
Concept Introduction:
Conversion-factor method is the one which can be used to convert one metric unit into another. With this, any unit can be converted to another unit by means of ratio. The ratio that is used to convert unit is known as conversion-factor. For example, one kilogram can be converted into gram by multiplying with 1000.
(d)
Expert Solution
Answer to Problem 1.133QP
The given millimeters is converted to centimeters as 3.71 cm.
Explanation of Solution
Given distance in problem statement is 37.1 mm.
Since, 1mm=10-3m and 1cm=10-2m, we can use this to convert the above unit in centimeters,
37.1 mm × 10-3 m1 mm × 1 cm10-2 m=3.71 cm
The given distance was converted into centimeters as shown above.
Want to see more full solutions like this?
Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell