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)
The Concept of Aromaticity
21.15 State the number of 2p orbital electrons in each molecule or ion.
(a)
(b)
(e)
(f)
(c)
(d)
(h)
(i)
DA
(k)
21.16 Which of the molecules and ions given in Problem 21.15 are aromatic according to the
Hückel criteria? Which, if planar, would be antiaromatic?
21.17 Which of the following structures are considered aromatic according to the Hückel
criteria?
---0-0
(a)
(b)
(c)
(d)
(e)
(h)
H
-H
.8.0-
21.18 Which of the molecules and ions from Problem 21.17 have electrons donated by a
heteroatom?
1. Show the steps necessary to make 2-methyl-4-nonene using a
Wittig reaction. Start with triphenylphosphine and an alkyl
halide. After that you may use any other organic or inorganic
reagents.
2. Write in the product of this reaction:
CH3
CH₂
(C6H5)₂CuLi
H₂O+
3. Name this compound properly, including stereochemistry.
H₂C
H3C
CH3
OH
4. Show the step(s) necessary to transform the compound on the
left into the acid on the right.
Bri
CH2
5. Write in the product of this
LiAlH4
Br
H₂C
OH
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.