cm3. A piece of aluminum has a mass of 75.2 grams. What is the volume (in ml) for the (Note: The density of aluminum is 2.70 g/cm.) 75.29 2.10g /cms (density. bensing= Mass muss ns.2 Volume = pesity Volume The density of cooking oil is 0.92 g/mL. What is the mass for 250 deciliters of the oi 210 = nensity xVolume. 5pd <2メ 100oml IL 10o0mL. 15000ml 25000 md x C.92 4 = 23000ml ニ Complete each of the following temperature conversions. (You must show your work fo conversion.) Please look up necessary equations in your book. a. 94 °F = ? °C 34.444 °c (94°F - 32) x 5/9=34.444°c b. 118.25 K = ? °C -154.9°C 1l8.25K-2n315 = -154.9°c %3D
cm3. A piece of aluminum has a mass of 75.2 grams. What is the volume (in ml) for the (Note: The density of aluminum is 2.70 g/cm.) 75.29 2.10g /cms (density. bensing= Mass muss ns.2 Volume = pesity Volume The density of cooking oil is 0.92 g/mL. What is the mass for 250 deciliters of the oi 210 = nensity xVolume. 5pd <2メ 100oml IL 10o0mL. 15000ml 25000 md x C.92 4 = 23000ml ニ Complete each of the following temperature conversions. (You must show your work fo conversion.) Please look up necessary equations in your book. a. 94 °F = ? °C 34.444 °c (94°F - 32) x 5/9=34.444°c b. 118.25 K = ? °C -154.9°C 1l8.25K-2n315 = -154.9°c %3D
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:り。
cm3.
A piece of aluminum has a mass of 75.2 grams. What is the volume (in ml) for the p
(Note: The density of aluminum is 2.70 g/cm³.)
2.10g /cms (dens;ty.
bensiig = Mass
muss
ns.2
Volume =
pesity
Volume
The density of cooking oil is 0.92 g/mL. What is the mass for 250 deciliters of the oil?
210
= nensity *Volume.
IL
5pd 22メ
100omd
= 25000ml
IL
10o0mL.
25000md
25000 md x
C.92 4
= 23000ml
ニ
Complete each of the following temperature conversions. (You must show your work for
conversion.) Please look up necessary equations in your book.
a. 94 °F = ? °C
34.444 °c
(94°F - 32) x 5/9=34.444°C
b. 118.25 K = ? °C -154.9°c
118.25K-2n315= -154.9°c
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