At 4.0 ° C , pure water has a density of 1.00 g / mL . At 60.0 ° C , the density is 0.98 g / mL . Calculate the volume in mL of 1.00 g of water at each temperature, and then calculate the percentage increase in volume that occurs as water is heated from 4.0 ° C to 60.0 ° C .
At 4.0 ° C , pure water has a density of 1.00 g / mL . At 60.0 ° C , the density is 0.98 g / mL . Calculate the volume in mL of 1.00 g of water at each temperature, and then calculate the percentage increase in volume that occurs as water is heated from 4.0 ° C to 60.0 ° C .
Solution Summary: The author explains that the density of any substance is the mass of that substance divided by its volume.
At
4.0
°
C
, pure water has a density of
1.00
g
/
mL
. At
60.0
°
C
, the density is
0.98
g
/
mL
. Calculate the volume in
mL
of
1.00
g
of water at each temperature, and then calculate the percentage increase in volume that occurs as water is heated from
4.0
°
C
to
60.0
°
C
.
When 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol.
When 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol.
Which experimental number must be initialled by the Lab TA for the first run of Part 1 of the experiment?
a) the heat capacity of the calorimeter
b) Mass of sample
c) Ti
d) The molarity of the HCl
e) Tf
Predict products for the Following organic rxn/s by
writing the structurels of the correct products. Write
above the line provided"
your answer
D2
①CH3(CH2) 5 CH3 + D₂ (adequate)"
+
2
mited)
19
Spark
Spark
por every item.
4 CH 3 11
3 CH 3 (CH2) 4 C-H + CH3OH
CH2 CH3 + CH3 CH2OH
0
CH3
fou
+
KMnDy→
C43
+ 2 KMn Dy→→
C-OH
")
0
C-OH
1110
(4.)
9+3
=C
CH3
+ HNO 3
0
+ Heat>
+ CH3 C-OH + Heat
CH2CH3
- 3
2
+ D Heat H
3
CH 3 CH₂ CH₂ C = CH + 2 H₂ →
2
2
Chapter 1 Solutions
Chemistry for Today: General Organic and Biochemistry
General, Organic, and Biological Chemistry - 4th edition
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The Creation of Chemistry - The Fundamental Laws: Crash Course Chemistry #3; Author: Crash Course;https://www.youtube.com/watch?v=QiiyvzZBKT8;License: Standard YouTube License, CC-BY