The picture has to be draw for the given explanation. Concept Introduction: Osmotic pressure: Osmosis can be prevented by applying a pressure to the solution. The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution. Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution. The osmotic pressure can be given by the equation, Π=MRT Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature
The picture has to be draw for the given explanation. Concept Introduction: Osmotic pressure: Osmosis can be prevented by applying a pressure to the solution. The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution. Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution. The osmotic pressure can be given by the equation, Π=MRT Here,Π=Osmotic pressure M=Molarity of solution R= Gas law constant T=Temperature
Solution Summary: The author explains that osmosis can be prevented by applying a pressure to the solution.
Interpretation: The picture has to be draw for the given explanation.
Concept Introduction:
Osmotic pressure: Osmosis can be prevented by applying a pressure to the solution. The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution.
Colligative properties of a substance include the depression in the freezing point, elevation of boiling-point and osmotic pressure. These are dependant only on the number present and not based on the solute particles present in an ideal solution.
The osmotic pressure can be given by the equation,
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 11 Solutions
WebAssign for Zumdahl/Zumdahl/DeCoste's Chemistry, 10th Edition [Instant Access], Single-Term