CHEMISTRY:SCI.IN CONTEXT (CL)-PACKAGE
5th Edition
ISBN: 9780393628173
Author: Gilbert
Publisher: NORTON
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Chapter 18.4, Problem 4PE
Interpretation Introduction
Interpretation: The value of
Concept introduction: Oxidation half-cell reaction (loss of electrons) takes place at anode. Reduction half reaction (gain of electrons) takes place at cathode.
To determine: The value of
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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
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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 18 Solutions
CHEMISTRY:SCI.IN CONTEXT (CL)-PACKAGE
Ch. 18.2 - Prob. 1PECh. 18.2 - Prob. 2PECh. 18.3 - Prob. 3PECh. 18.4 - Prob. 4PECh. 18.6 - Prob. 5PECh. 18.6 - Prob. 6PECh. 18.7 - Prob. 7PECh. 18.9 - Prob. 8PECh. 18 - Prob. 18.1VPCh. 18 - Prob. 18.2VP
Ch. 18 - Prob. 18.3VPCh. 18 - Prob. 18.4VPCh. 18 - Prob. 18.5VPCh. 18 - Prob. 18.6VPCh. 18 - Prob. 18.7VPCh. 18 - Prob. 18.8VPCh. 18 - Prob. 18.9VPCh. 18 - Prob. 18.10VPCh. 18 - Prob. 18.11QPCh. 18 - Prob. 18.12QPCh. 18 - Prob. 18.13QPCh. 18 - Prob. 18.14QPCh. 18 - Prob. 18.15QPCh. 18 - Prob. 18.16QPCh. 18 - Prob. 18.17QPCh. 18 - Prob. 18.18QPCh. 18 - Prob. 18.19QPCh. 18 - Prob. 18.20QPCh. 18 - Prob. 18.21QPCh. 18 - Prob. 18.22QPCh. 18 - Prob. 18.23QPCh. 18 - Prob. 18.24QPCh. 18 - Prob. 18.25QPCh. 18 - Prob. 18.26QPCh. 18 - Prob. 18.27QPCh. 18 - Prob. 18.28QPCh. 18 - Prob. 18.29QPCh. 18 - Prob. 18.30QPCh. 18 - Prob. 18.31QPCh. 18 - Prob. 18.32QPCh. 18 - Prob. 18.33QPCh. 18 - Prob. 18.34QPCh. 18 - Prob. 18.35QPCh. 18 - Prob. 18.36QPCh. 18 - Prob. 18.37QPCh. 18 - Prob. 18.38QPCh. 18 - Prob. 18.39QPCh. 18 - Prob. 18.40QPCh. 18 - Prob. 18.41QPCh. 18 - Prob. 18.42QPCh. 18 - Prob. 18.43QPCh. 18 - Prob. 18.44QPCh. 18 - Prob. 18.45QPCh. 18 - Prob. 18.46QPCh. 18 - Prob. 18.47QPCh. 18 - Prob. 18.48QPCh. 18 - Prob. 18.49QPCh. 18 - Prob. 18.50QPCh. 18 - Prob. 18.51QPCh. 18 - Prob. 18.52QPCh. 18 - Prob. 18.53QPCh. 18 - Prob. 18.54QPCh. 18 - Prob. 18.55QPCh. 18 - Prob. 18.56QPCh. 18 - Prob. 18.57QPCh. 18 - Prob. 18.58QPCh. 18 - Prob. 18.59QPCh. 18 - Prob. 18.60QPCh. 18 - Prob. 18.61QPCh. 18 - Prob. 18.62QPCh. 18 - Prob. 18.63QPCh. 18 - Prob. 18.64QPCh. 18 - Prob. 18.65QPCh. 18 - Prob. 18.66QPCh. 18 - Prob. 18.67QPCh. 18 - Prob. 18.68QPCh. 18 - Prob. 18.69QPCh. 18 - Prob. 18.70QPCh. 18 - Prob. 18.71QPCh. 18 - Prob. 18.72QPCh. 18 - Prob. 18.73QPCh. 18 - Prob. 18.74QPCh. 18 - Prob. 18.75QPCh. 18 - Prob. 18.76QPCh. 18 - Prob. 18.77QPCh. 18 - Prob. 18.78QPCh. 18 - Prob. 18.79QPCh. 18 - Prob. 18.80QPCh. 18 - Prob. 18.81QPCh. 18 - Prob. 18.82QPCh. 18 - Prob. 18.83QPCh. 18 - Prob. 18.84QPCh. 18 - Prob. 18.85QPCh. 18 - Prob. 18.86QPCh. 18 - Prob. 18.87QPCh. 18 - Prob. 18.88QPCh. 18 - Prob. 18.89QPCh. 18 - Prob. 18.90QPCh. 18 - Prob. 18.91APCh. 18 - Prob. 18.92APCh. 18 - Prob. 18.93APCh. 18 - Prob. 18.94APCh. 18 - Prob. 18.95APCh. 18 - Prob. 18.96APCh. 18 - Prob. 18.97APCh. 18 - Prob. 18.98APCh. 18 - Prob. 18.99APCh. 18 - Prob. 18.100AP
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- 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.arrow_forwardQ6: Using acetic acid as the acid, write the balanced chemical equation for the protonation of the two bases shown (on the -NH2). Include curved arrows to show the mechanism. O₂N- O₂N. -NH2 -NH2 a) Which of the two Bronsted bases above is the stronger base? Why? b) Identify the conjugate acids and conjugate bases for the reactants. c) Identify the Lewis acids and bases in the reactions.arrow_forwardQ5: For the two reactions below: a) Use curved electron-pushing arrows to show the mechanism for the reaction in the forward direction. Redraw the compounds to explicitly illustrate all bonds that are broken and all bonds that are formed. b) Label Bronsted acids and bases in the left side of the reactions. c) For reaction A, which anionic species is the weakest base? Which neutral compound is the stronger acid? Is the forward or reverse reaction favored? d) Label Lewis acids and bases, nucleophiles and electrophiles in the left side of the reactions. A. 용 CH3OH я хон CH3O OH B. HBr CH3ONa NaBr CH3OHarrow_forward
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