The number of moles and number of ions of each type in 88 mL of 1.75 M magnesium chloride 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 liter 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 moles of ions is as follows: moles of ion of compound ( mol ) = [ ( moles of compound ( mol ) ) ( total moles of ion ( mol ) 1 mole of compound ) ] The expression to calculate the number of ions is as follows: number of ions = ( moles of ions ( mol ) ) ( 6 .022 × 10 23 ions 1 mole of ions )
The number of moles and number of ions of each type in 88 mL of 1.75 M magnesium chloride 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 liter 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 moles of ions is as follows: moles of ion of compound ( mol ) = [ ( moles of compound ( mol ) ) ( total moles of ion ( mol ) 1 mole of compound ) ] The expression to calculate the number of ions is as follows: number of ions = ( moles of ions ( mol ) ) ( 6 .022 × 10 23 ions 1 mole of ions )
The number of moles and number of ions of each type in 88 mL of 1.75M magnesium chloride 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 liter 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 moles of ions is as follows:
moles ofion of compound(mol)=[(moles of compound(mol))(total 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)
(b)
Interpretation Introduction
Interpretation:
The number of moles and number of ions of each type in 321 mL of a solution containing 0.22 g aluminium sulfate per liter 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 liter of solution. Unit of molarity is mol/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))]
The expression to calculate the moles of ions is as follows:
moles ofion of compound(mol)=[(moles of compound(mol))(total 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)
(c)
Interpretation Introduction
Interpretation:
The number of moles and number of ions of each type in 1.65 mL of a solution containing 8.83×1021 formula units of cesium nitrate per liter is to be calculated.
Concept introduction:
A formula unit is used for the ionic compound to represent their empirical formula. The expression to calculate the moles of a compound when the volume of solution and formula unit of a compound is given is as follows:
moles of a compound(mol)=[(volume of solution(L))(given formula unit of compound(FU))(1 mole of compound6.022×1023FU)]
The expression to calculate the moles of ions is as follows:
moles ofion of compound(mol)=[(moles of compound(mol))(total 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)
Can the target compound at right be efficiently synthesized in good yield from the unsubstituted benzene at left?
?
starting
material
target
If so, draw a synthesis below. If no synthesis using reagents ALEKS recognizes is possible, check the box under the drawing area.
Be sure you follow the standard ALEKS rules for submitting syntheses.
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Note for advanced students: you may assume that you are using a large excess of benzene as your starting material.
C
:0
T
Add/Remove step
G
The following equations represent the formation of compound MX. What is the AH for the
electron affinity of X (g)?
X₂ (g) → 2X (g)
M (s) → M (g)
M (g)
M (g) + e-
AH = 60 kJ/mol
AH = 22 kJ/mol
X (g) + e-X (g)
M* (g) +X (g) → MX (s)
AH = 118 kJ/mol
AH = ?
AH = -190 kJ/mol
AH = -100 kJ/mol
a)
-80 kJ
b)
-30 kJ
c)
-20 kJ
d)
20 kJ
e)
156 kJ
A covalent bond is the result of the
a)
b)
c)
d)
e)
overlap of two half-filled s orbitals
overlap of a half-filled s orbital and a half-filled p orbital
overlap of two half-filled p orbitals along their axes
parallel overlap of two half-filled parallel p orbitals
all of the above
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