The volume (L) of 2.050 M copper ( II ) nitrate that must be diluted with water to prepare 750.0 mL of a 0.8543 M solution is to be calculated. Concept introduction: Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed. The expression to relate the molarity of a concentrated and dilute solution is: M 1 V 1 = M 2 V 2 (1) Here, M 1 is the molarity of the dilute solution. V 1 is the volume of dilute solution. M 2 is the molarity of the concentrated solution. V 2 is the volume of the concentrated solution.
The volume (L) of 2.050 M copper ( II ) nitrate that must be diluted with water to prepare 750.0 mL of a 0.8543 M solution is to be calculated. Concept introduction: Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed. The expression to relate the molarity of a concentrated and dilute solution is: M 1 V 1 = M 2 V 2 (1) Here, M 1 is the molarity of the dilute solution. V 1 is the volume of dilute solution. M 2 is the molarity of the concentrated solution. V 2 is the volume of the concentrated solution.
The volume (L) of 2.050Mcopper(II)nitrate that must be diluted with water to prepare 750.0 mL of a 0.8543M solution is to be calculated.
Concept introduction:
Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.
The expression to relate the molarity of a concentrated and dilute solution is:
M1V1=M2V2 (1)
Here,
M1 is the molarity of the dilute solution.
V1 is the volume of dilute solution.
M2 is the molarity of the concentrated solution.
V2 is the volume of the concentrated solution.
(b)
Interpretation Introduction
Interpretation:
Volume (L) of 1.63M calcium chloride that must be diluted with water to prepare 350 mL of a 2.86×10−2M chloride ion solution is to be calculated.
Concept introduction:
Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.
The expression to relate the molarity of a concentrated and dilute solution is:
M1V1=M2V2 (1)
Here,
M1 is the molarity of the dilute solution.
V1 is the volume of dilute solution.
M2 is the molarity of the concentrated solution.
V2 is the volume of the concentrated solution.
(c)
Interpretation Introduction
Interpretation:
The final volume (L) of a 0.0700M solution prepared by diluting 18.0 mL of 0.155M lithium carbonate with water is to be calculated.
Concept introduction:
Dilution is the process of converting a concentrated solution into a dilute solution by adding the solvent. In the resultant solution, the amount of solute remains fixed but the volume of the solution increases. The moles of solute before and after dilution remain fixed.
The expression to relate the molarity of a concentrated and dilute solution is:
The sum of the numbers in the name of isA. 11; B. 13; C. 10; D. 12; E. none of the other answers iscorrect. I believe the awnser should be E to this problem but the solution to this problem is D 12. I'm honestly unsure how that's the solution. If you can please explain the steps to this type of problem and how to approach a problem like this it would be greatly appreciated!
Consider the following data for phosphorus:
g
atomic mass
30.974
mol
electronegativity
2.19
kJ
electron affinity
72.
mol
kJ
ionization energy
1011.8
mol
kJ
heat of fusion
0.64
mol
You may find additional useful data in the ALEKS Data tab.
Does the following reaction absorb or release energy?
2+
+
(1) P (g) + e
→ P (g)
Is it possible to calculate the amount of
energy absorbed
or released by reaction (1) using only the data above?
If you answered yes to the previous question, enter the
amount of energy absorbed or released by reaction (1):
Does the following reaction absorb or release energy?
00
release
absorb
Can't be decided with the data given.
yes
no
☐ kJ/mol
(²) P* (8) +
+
+ e →>>
P (g)
Is it possible to calculate the amount of energy absorbed
or released by reaction (2) using only the data above?
If you answered yes to the previous question, enter the
amount of energy absorbed or released by reaction (2):
☐
release
absorb
Can't be decided with the data given.
yes
no
kJ/mol
а
The number of hydrogens in an alkyne that has a main chain of 14carbons to which are attached a cyclobutyl ring, a benzene ring, an–OH group, and a Br is A. 34; B. 35; C. 36; D. 24; E. 43
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