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.
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5. A solution of sucrose is fermented in a vessel until the evolution of CO2 ceases. Then, the
product solution is analyzed and found to contain, 45% ethanol; 5% acetic acid; and 15%
glycerin by weight.
If the original charge is 500 kg, evaluate;
e. The ratio of sucrose to water in the original charge (wt/wt).
f. Moles of CO2 evolved.
g. Maximum possible amount of ethanol that could be formed.
h. Conversion efficiency.
i. Per cent excess of excess reactant.
Reactions:
Inversion reaction: C12H22O11 + H2O →2C6H12O6
Fermentation reaction: C6H12O6 →→2C2H5OH + 2CO2
Formation of acetic acid and glycerin: C6H12O6 + C2H5OH + H₂O→ CH3COOH + 2C3H8O3
Show work. don't give Ai generated solution. How many carbons and hydrogens are in the structure?
13. (11pts total) Consider the arrows pointing at three different carbon-carbon bonds in the
molecule depicted below.
Bond B
2°C. +2°C. cleavage
Bond A
•CH3 + 26.← Cleavage
2°C. +
Bond C
+3°C•
CH3 2C
Cleavage
E
2°C. 26.
weakest bond
Intact molecule
Strongest 3°C 20.
Gund
Largest
argest
a. (2pts) Which bond between A-C is weakest? Which is strongest? Place answers in
appropriate boxes.
C
Weakest
bond
A
Produces
Most
Bond
Strongest
Bond
Strongest Gund
produces least stable
radicals
Weakest
Stable radical
b. (4pts) Consider the relative stability of all cleavage products that form when bonds A,
B, AND C are homolytically cleaved/broken. Hint: cleavage products of bonds A, B,
and C are all carbon radicals.
i. Which ONE cleavage product is the most stable? A condensed or bond line
representation is fine.
13°C. formed in
bound C
cleavage
ii. Which ONE cleavage product is the least stable? A condensed or bond line
representation is fine.
• CH3
methyl radical
Formed in Gund A Cleavage
c.…
Chapter 1 Solutions
OWLv2 for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 1 term (6 months)
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