The mass ( g ) of solute needed to make 475 mL of 5.62 × 10 − 2 M potassium sulphate is to be calculated. Concept introduction: Molarity ( M ) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L . The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows: Moles of compound ( mol ) = [ volume of solution ( L ) ( molarity of solution ( mol ) 1L of solution ) ] The expression to calculate the amount of compound when moles of the compound and molecular mass are given: amount of compound ( g ) = moles of compound ( mol ) ( molecular mass of compound ( g ) 1 mole of compound )
The mass ( g ) of solute needed to make 475 mL of 5.62 × 10 − 2 M potassium sulphate is to be calculated. Concept introduction: Molarity ( M ) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L . The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows: Moles of compound ( mol ) = [ volume of solution ( L ) ( molarity of solution ( mol ) 1L of solution ) ] The expression to calculate the amount of compound when moles of the compound and molecular mass are given: amount of compound ( g ) = moles of compound ( mol ) ( molecular mass of compound ( g ) 1 mole of compound )
The mass (g) of solute needed to make 475 mL of 5.62×10−2M potassium sulphate is to be calculated.
Concept introduction:
Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.
The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows:
Moles of compound(mol)=[volume of solution(L)(molarityofsolution(mol)1L of solution)]
The expression to calculate the amount of compound when moles of the compound and molecular mass are given:
amount of compound(g)=moles of compound(mol)(molecular mass of compound(g)1mole of compound)
(b)
Interpretation Introduction
Interpretation:
Molarity of a solution that contains 7.25 mg of calcium chloride in each millilitre is to be calculated.
Concept introduction:
Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.
The expression to calculate the molarity of a solution when moles of solute and volume of solution are given is as follows:
Molarity of solution(M)=moles of solute(mol)volume of solution(L)
The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:
Moles of compound(mol)=[given massof compound(g)(1moleof compound(mol)molecular mass of compound(g))]
(c)
Interpretation Introduction
Interpretation:
The number of Mg2+ ions in each milliliters of 0.184M magnesium bromide is to be calculated.
Concept introduction:
Molarity (M) is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is mol/L.
The expression to calculate the moles of the compound when molarity of solution and volume of solution are given is as follows:
Moles of compound(mol)=[volume of solution(L)(molarityofsolution(mol)1L of solution)]
The expression to calculate the amount of ions in moles is as follows:
amountofion(mol)=(moles of compound(mol))(moles of ion(mol)1mole of compound)
The expression to calculate the number of ions is as follows:
numberof ions=(moles of ions(mol))(6.022×1023ions1mole of ions)
Predict the major organic product(s) of the following reactions. Indicate which of the following mechanisms is in operation: SN1, SN2, E1, or E2.
(c)
(4pts)
Mechanism:
heat
(E1)
CH3OH
+
1.5pts each
_E1 _ (1pt)
Br
CH3OH
(d)
(4pts)
Mechanism:
SN1
(1pt)
(e)
(3pts)
1111 I
H
10
Ill!!
H
LDA
THF (solvent)
Mechanism: E2
(1pt)
NC
(f)
Bri!!!!!
CH3
NaCN
(3pts)
acetone
Mechanism: SN2
(1pt)
(SN1)
-OCH3
OCH3
1.5pts each
2pts for either product
1pt if incorrect
stereochemistry
H
Br
(g)
“,、
(3pts)
H
CH3OH
+21
Mechanism:
SN2
(1pt)
H
CH3
2pts
1pt if incorrect
stereochemistry
H
2pts
1pt if incorrect
stereochemistry
A mixture of butyl acrylate and 4'-chloropropiophenone has been taken for proton NMR analysis. Based on this proton NMR, determine the relative percentage of each compound in the mixture
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