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Fundamentals of Analytical Chemistry: 8 ed.
Chapter 4
Chapter 4
4-1
(a) The millimole is an amount of a chemical species, such as an atom, an ion, a molecule
or an electron. A millimole millimol e contains 6.02 × 10
23
particles
∗
mole
10
mole
−3
millimole
=
6.02 × 10
20
particles millimole
(b) The molar mass is the mass in grams of one mole of a chemical species. (c) The millimolar mass is the mass in grams of one millimole of a chemical species. (d) Parts per million, c ppm, is a term expressing the concentration of very dilute solutions.
Thus, cppm =
mass of solute mass of solution
×
6
10 ppm
The units of mass in the numerator and the denominator must be the same.
4-2
The species molarity of a solution expresses the equilibrium concentration of a chemical species in terms of moles per liter. The analytical molarity of a solution gives the total number of moles of a solute in one liter. The species molarity takes into account chemical reactions that occur in solution. The analytical molarity specifies how the solution was prepared, but does not account for any subsequent reactions.
4-3
1L =
1000 mL
1M =
×
1L 1 mole L
×
m × mL 100 cm
1 cm 3
L 10 −3 m 3
=
3
1 mole 10 −3 m 3
=
10 −3 m 3
th
Fundamentals of Analytical Chemistry: 8 ed.
4-4
(a) 3.2 × 105 Hz
kHz
×
−8
(b) 4.56 × 10 g
10 9 ng
×
6
(d) 6.5 × 10 s
4
4-5
×
5.43 g Na 3 PO 4 ×
6.5 Ms
=
6
10 s
µm
×
1000 g
72 kg
=
1 mol Na 3 PO 4 163.94g
4-6
6.76 mol K 3 PO 4 ×
4-7
(a) 4.96 g B2 O 3 ×
89.6 µm
=
3
10 nm
kg
843 mmol
=
3 10 µmol
Ms
(e) 8.96 × 10 nm
(f) 72,000g
mmol
×
×
45.6 ng
=
g
(c) 8.43 × 10 5 µmol
320 kHz
=
1000 Hz
Chapter 4
3 mol K + mol K 3 PO 4
×
69.62 g B2 O 3
mol Na 3 PO 4
6.02 × 10 23 K +
×
mol B2 O 3
23
3 mol Na +
mol K
+
×
6.02 × 10 Na + mol Na
+
25
= 1.22 × 10 K
= 5.98 × 10
22
+
= 0.0712 mol B2 O 3
(b)
333 mg Na 2 B4 O 7 • 10 H 2 O × = 8.73 × 10
−4
g 1000 mg
×
mol Na 2 B4 O 7 ⋅ 10H 2 O 381.37g
mol Na 2 B4 O 7 ⋅ 10H 2 O
(c) 8.75 g Mn 3O 4 ×
mol Mn 3O 4 228.81 g Mn 3O 4
(d) 167.2 mg CaC 2 O 4 ×
g 1000 mg
×
= 0.0382 mol Mn 3 O 4
mol CaC 2 O 4 128.10 g CaC 2 O 4
= 1.31 × 10
−3
mol CaC 2 O 4
Na +
th
Fundamentals of Analytical Chemistry: 8 ed.
4-8
(a) 57 mg P2 O 5 ×
g 1000 mg
(b) 12.92 g CO 2 ×
×
mol P2 O 5 141.94 g P2 O 5
mol CO 2
×
44.01 g CO 2
(c) 40.0 g NaHCO3 ×
Chapter 4
×
1000 mmol mol
mol NaHCO 3 84.01 g NaHCO 3
×
1000 mmol mol
= 0.40 mmol P2 O 5
= 293.6 mmol CO 2
1000 mmol mol
= 476 mmol NaHCO 3
(d)
850 mg MgNH 4 PO 4 ×
g 1000 mg
×
mol MgNH 4 PO 4 137.32 g MgNH 4 PO 4
×
1000 mmol mol
= 6.2 mmol MgNH 4 PO 4
4-9
(a) 3
−3
3.25 × 10 M KMnO4 ≡
3.25 × 10 − mol KMnO 4 L
×
1000 mmol mol
× 2.00 L
= 6.50 mmol KMnO 4
(b)
0.0555 M KSCN ≡
0.0555 mol KSCN L
×
1000 mmol mol
×
L 1000 mL
× 750 mL
= 41.6 mmol KSCN
(c)
5.41 ppm CuSO4 ≡
5.41 mg CuSO4 L ×
(d) 0.333 M KCl ≡
g 1000 mg
L 1000 mL
0.333 mol KCl L
×
×
×
mol CuSO4 159.61 g CuSO4
× 250 mL = 8.47 × 10
1000 mmol mol
−3
×
1000 mmol mol
mmol CuSO4
× 3.50 L = 1165.6 mmol KCl
th
Fundamentals of Analytical Chemistry: 8 ed. 4-10
Chapter 4
(a)
0.320 M HClO 4 ≡
0.320 mol HClO 4
×
L
1000 mmol mol
×
L 1000 mL
× 175 mL
= 56.0 mmol HClO4
(b) −3
−3
8.05 × 10 M K 2 CrO 4 ≡
8.05 × 10 mol K 2 CrO 4
×
L
1000 mmol mol
× 15.0 L
= 121 mmol K 2 CrO 4
(c)
6.75 ppm AgNO 3 ≡
6.75 mg AgNO 3
×
L
g 1000 mg
×
mol AgNO 3 169.87 g AgNO 3
×
1000 mmol mol
×
5.00 L = 0.199 mmol AgNO 3 (d)
0.0200 M KOH ≡
0.0200 mol KOH L
×
1000 mmol mol
×
L 1000 mL
× 851 mL
= 17.0 mmol KOH
4-11
(a) 0.777 mol HNO 3 ×
(b) 500 mmol MgO ×
63.01 g HNO 3 mol HNO3 mol
1000 mmol
(c) 22.5 mol NH 4 NO 3 ×
×
×
1000 mg g
40.30 g MgO mol MgO
80.04 g NH 4 NO 3 mol NH 4 NO 3
×
4
= 4.90 × 10 mg HNO 3
×
1000 mg
1000 mg g
g
4
= 2.015 × 10 mg MgO
6
= 1.80 × 10 mg NH 4 NO 3
(d)
4.32 mol ( NH 4 ) 2 Ce( NO 3 ) 6 × 6
548.23 g ( NH 4 ) 2 Ce( NO 3 ) 6 mol ( NH 4 ) 2 Ce( NO 3 ) 6
= 2.37 × 10 mg ( NH 4 ) 2 Ce( NO 3 ) 6
×
1000 mg g
th
Fundamentals of Analytical Chemistry: 8 ed.
4-12
(a) 7.1 mol KBr ×
119.0 g KBr mol KBr
(b) 20.1 mmol PbO ×
= 840 g KBr
mol 1000 mmol
(c) 3.76 mol MgSO 4 ×
Chapter 4
×
223.20 g PbO mol PbO
120.37 g MgSO 4
= 4.49 g PbO
= 452 g MgSO 4
mol MgSO 4
(d)
9.6 mmol Fe( NH 4 ) 2 (SO 4 ) 2 ⋅ 6H 2 O ×
mol 1000 mmol
×
392.23 g Fe( NH 4 ) 2 (SO 4 ) 2 ⋅ 6H 2 O mol Fe( NH 4 ) 2 (SO 4 ) 2 ⋅ 6H 2 O