The following equation can be used to relate the density of Equid water to Celsius temperature in the range from 0 °C to about 20 °C: d ( g / c m 3 ) = 0.99984 + ( 1.6945 × 10 − 2 t ) − ( 7.987 × 10 − 6 t 2 ) 1 + ( 1.6880 + 10 − 2 t ) a. To four significant figures, determine the density of water at 10 °C. b. At what temperature does water have a density of 0.99860 g/cm 3 ? c. In the following ways, show that the density passes through a maximum somewhere in the temperature range to which the equation applies. i. by estimation ii. by a graphical method iii. by a method based on differential calculus
The following equation can be used to relate the density of Equid water to Celsius temperature in the range from 0 °C to about 20 °C: d ( g / c m 3 ) = 0.99984 + ( 1.6945 × 10 − 2 t ) − ( 7.987 × 10 − 6 t 2 ) 1 + ( 1.6880 + 10 − 2 t ) a. To four significant figures, determine the density of water at 10 °C. b. At what temperature does water have a density of 0.99860 g/cm 3 ? c. In the following ways, show that the density passes through a maximum somewhere in the temperature range to which the equation applies. i. by estimation ii. by a graphical method iii. by a method based on differential calculus
Solution Summary: The author explains how the density of water is calculated using the following formula: d= mV Here, m is mass and V is volume.
The following equation can be used to relate the density of Equid water to Celsius temperature in the range from 0 °C to about 20 °C:
d
(
g
/
c
m
3
)
=
0.99984
+
(
1.6945
×
10
−
2
t
)
−
(
7.987
×
10
−
6
t
2
)
1
+
(
1.6880
+
10
−
2
t
)
a. To four significant figures, determine the density of water at 10 °C. b. At what temperature does water have a density of 0.99860 g/cm3? c. In the following ways, show that the density passes through a maximum somewhere in the temperature range to which the equation applies. i. by estimation ii. by a graphical method iii. by a method based on differential calculus
Explain how you could experimentally determine whether the outside temperature is higher or lower than 0 °C (32 °F) without using a thermometer.
The molar volumes of pure acetone and pure chloroform at room temperature are 72.0 cm^3 and 81.0 cm^3, respectively. The volume of a liquid mixture containing 0.6 moles of acetone and 0.5 moles of chloroform is
You are asked to calibrate a 25.00 mL volumetric pipet. You determine the temperature of your distilled water is exactly 30 degrees celsius. You carefully determined the mass of a clean, dry beaker and found it was 60.1703 g. You pulled the water up to the mark and transferred this to the beaker and found that the new mass was 85.2263 g. What is the actualy volume of the pipet?
I tried answering the question above by doing 85.2263 - 60.1703 = 25.056 g
I then used 25.056 and multipled it with .995646 (which is the density of water at 30 degrees celsius I believe)
The answer I got is 24.947
I am not sure if I am just doing the problem wrong or if it is simply my signifcant figures that are off? Please help :))
Chapter 1 Solutions
General Chemistry: Principles and Modern Applications, Loose Leaf Version (11th Edition)
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