(a) Interpretation: The number of atoms present in the sample of gold should be calculated. Concept Introduction: Moles of an element are calculated by dividing the given mass by its molar mass. Moles of an element = Mass MolarMass According to Avogadro’s number , the number of atoms present in one mole of a monoatomic element is 6.023 × 10 23 .
(a) Interpretation: The number of atoms present in the sample of gold should be calculated. Concept Introduction: Moles of an element are calculated by dividing the given mass by its molar mass. Moles of an element = Mass MolarMass According to Avogadro’s number , the number of atoms present in one mole of a monoatomic element is 6.023 × 10 23 .
Definition Definition Number of atoms/molecules present in one mole of any substance. Avogadro's number is a constant. Its value is 6.02214076 × 10 23 per mole.
Chapter 8, Problem 109AP
Interpretation Introduction
(a)
Interpretation:
The number of atoms present in the sample of gold should be calculated.
Concept Introduction:
Moles of an element are calculated by dividing the given mass by its molar mass.
Moles of an element =MassMolarMass
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.
Expert Solution
Answer to Problem 109AP
9.03×1021 atoms are present in 2.89g of gold.
Explanation of Solution
Mass of gold = 2.89g
Molar mass of gold = 196.96 g/mol
Moles of gold =MassMolarMass=2.89 g196.96 g/mol=0.015 mol
Number of atoms present in 0.015 mole of gold (Au)=(0.015×6.023×1023)=9.03×1021 atoms.
Interpretation Introduction
(b)
Interpretation:
The number of atoms present in the sample of platinum should be calculated.
Concept Introduction:
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.
Expert Solution
Answer to Problem 109AP
1.56×1020 atoms are present in 0.000259 mole of platinum.
Explanation of Solution
Moles of platinum present in the sample = 0.000259
Number of atoms present in 0.000259 mole of platinum (Pt)=(0.000259×6.023×1023)=1.56×1020.
Interpretation Introduction
(c)
Interpretation:
The number of atoms present in the sample of platinum should be calculated.
Concept Introduction:
Moles of an element are calculated by dividing the given mass by its molar mass.
Moles of an element =MassMolarMass
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.
Expert Solution
Answer to Problem 109AP
8.01×1017 atoms are present in 0.000259 g of platinum.
Explanation of Solution
Mass of platinum = 0.000259g
Molar mass of platinum = 195.08g/mol
Moles of platinum =MassMolarMass=0.000259 g195.08 g/mol=1.33×10−6 mol
Number of atoms present in 1.33×10−6 mole of platinum (Pt)=(1.33×10−6×6.023×1023)=8.01×1017.
Interpretation Introduction
(d)
Interpretation:
The number of atoms present in the sample of magnesium should be calculated.
Concept Introduction:
Moles of an element are calculated by dividing the given mass by its molar mass.
Moles of an element =MassMolarMass
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.
Expert Solution
Answer to Problem 109AP
2.25×1025 atoms are present in 2.0lb of platinum.
Explanation of Solution
Mass of magnesium = 2.0lb = 907.184g [Since, 1lb = 453.592g]
Molar mass of magnesium = 24.3g/mol
Moles of magnesium =MassMolarMass=907.184 g24.3 g/mol=37.33 mol
Number of atoms present in 37.33 moles of magnesium (Mg)=(37.33×6.023×1023)=2.25×1025.
Interpretation Introduction
(e)
Interpretation:
The number of atoms present in the sample of mercury should be calculated.
Concept Introduction:
Mass can be calculated by multiplying the density by the volume.
Mass=Density×Volume
Moles of an element are calculated by dividing the given mass by its molar mass.
Moles of an element =MassMolarMass
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.
Expert Solution
Answer to Problem 109AP
7.7×1022 atoms are present in 1.90 mL of mercury.
Explanation of Solution
Volume of mercury = 1.90mL
Density of mercury = 13.6g/mL
Mass of mercury =(Density×Volume)=(13.6g/mL×1.90mL)=25.84 g
Molar mass of mercury = 200.59g/mol
Moles of mercury =MassMolar Mass=25.84 g200.59 g/mol=0.128 mol
Number of atoms present in 37.33 moles of
Mercury (Hg)=(0.128×6.023×1023)=7.7×1022.
Interpretation Introduction
(f)
Interpretation:
The number of atoms present in the sample of tungsten should be calculated.
Concept Introduction:
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.
Expert Solution
Answer to Problem 109AP
2.58×1023 atoms are present in 4.30 moles of tungsten.
Explanation of Solution
Moles of tungsten = 4.30
Number of atoms present in 4.30 moles of
Tungsten (W)=(4.30×6.023×1023)=2.58×1023.
Interpretation Introduction
(f)
Interpretation:
The number of atoms present in the sample of tungsten should be calculated.
Concept Introduction:
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.
Expert Solution
Answer to Problem 109AP
2.58×1023 atoms are present in 4.30 moles of tungsten.
Explanation of Solution
Moles of tungsten = 4.30
Number of atoms present in 4.30 moles of
Tungsten (W)=(4.30×6.023×1023)=2.58×1023.
Interpretation Introduction
(g)
Interpretation:
The number of atoms present in the sample of tungsten should be calculated.
Concept Introduction:
Moles of an element are calculated by dividing the given mass by its molar mass.
Moles of an element =MassMolarMass
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is 6.023×1023.
Expert Solution
Answer to Problem 109AP
1.38×1022 atoms are present in 4.30g of tungsten.
Explanation of Solution
Mass of tungsten = 4.30g
Molar mass of tungsten = 183.84g/mol
Moles of tungsten =MassMolar Mass=4.30 g183.84 g/mol=0.023 mol
Number of atoms present in 0.023 moles of
Tungsten (W)=(0.023×6.023×1023)=1.38×1022.
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(f) SO:
Best Lewis Structure
3
e group geometry:_
shape/molecular geometry:,
(g) CF2CF2
Best Lewis Structure
polarity:
e group arrangement:_
shape/molecular geometry:
(h) (NH4)2SO4
Best Lewis Structure
polarity:
e group arrangement:
shape/molecular geometry:
polarity:
Sketch (with angles):
Sketch (with angles):
Sketch (with angles):
1.
Problem Set 3b
Chem 141
For each of the following compounds draw the BEST Lewis Structure then sketch the molecule (showing
bond angles). Identify (i) electron group geometry (ii) shape around EACH central atom (iii) whether the
molecule is polar or non-polar (iv)
(a) SeF4
Best Lewis Structure
e group arrangement:_
shape/molecular geometry:
polarity:
(b) AsOBr3
Best Lewis Structure
e group arrangement:_
shape/molecular geometry:
polarity:
Sketch (with angles):
Sketch (with angles):
(c) SOCI
Best Lewis Structure
2
e group arrangement:
shape/molecular geometry:_
(d) PCls
Best Lewis Structure
polarity:
e group geometry:_
shape/molecular geometry:_
(e) Ba(BrO2):
Best Lewis Structure
polarity:
e group arrangement:
shape/molecular geometry:
polarity:
Sketch (with angles):
Sketch (with angles):
Sketch (with angles):
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