Concept explainers
(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
According to
Answer to Problem 109AP
Explanation of Solution
Mass of gold = 2.89g
Molar mass of gold = 196.96 g/mol
Moles of gold
Number of atoms present in 0.015 mole of gold
(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
Answer to Problem 109AP
Explanation of Solution
Moles of platinum present in the sample = 0.000259
Number of atoms present in 0.000259 mole of platinum
(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
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is
Answer to Problem 109AP
Explanation of Solution
Mass of platinum = 0.000259g
Molar mass of platinum = 195.08g/mol
Moles of platinum
Number of atoms present in
(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
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is
Answer to Problem 109AP
Explanation of Solution
Mass of magnesium = 2.0lb = 907.184g [Since, 1lb = 453.592g]
Molar mass of magnesium = 24.3g/mol
Moles of magnesium
Number of atoms present in 37.33 moles of magnesium
(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.
Moles of an element are calculated by dividing the given mass by its molar mass.
Moles of an element
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is
Answer to Problem 109AP
Explanation of Solution
Volume of mercury = 1.90mL
Density of mercury = 13.6g/mL
Mass of mercury
Molar mass of mercury = 200.59g/mol
Moles of mercury
Number of atoms present in 37.33 moles of
Mercury
(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
Answer to Problem 109AP
Explanation of Solution
Moles of tungsten = 4.30
Number of atoms present in 4.30 moles of
Tungsten
(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
Answer to Problem 109AP
Explanation of Solution
Moles of tungsten = 4.30
Number of atoms present in 4.30 moles of
Tungsten
(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
According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is
Answer to Problem 109AP
Explanation of Solution
Mass of tungsten = 4.30g
Molar mass of tungsten = 183.84g/mol
Moles of tungsten
Number of atoms present in 0.023 moles of
Tungsten
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Chapter 8 Solutions
Introductory Chemistry: A Foundation 8th Edition Access Code
- Using the average atomic masses given inside the front cover of this book, calculate the mass in grams of each of the following samples. l type='a'> 2.17 moles ot germanium, Ge 4.24 mmcl of Iead(II) chloride (1 mmol = 1/1000 moI) 0.0971 mole of ammonia, NH3 :math>4.26103moles of hexane, C6H14 i>1.71 moles of iodine monochloride, ICIarrow_forwarditamin B12 , cyancobalamin, is essential for human nutrition. Its molecular formula is C63H88CoN14O14P . A lack of this vita min in the diet can lead to anemia. Cyanocohalamin is the form of the vitamin found in vitamin supplements. l type='a'> What is the molar mass of cyanocobalamin to two decimal places? How many moles of cyanocohalamin molecules are present in 250 mg of cyanocobalamin? What is the mass of 0.60 mole of cyanocobalamin? How many atoms of hydrogen are in 1 .0 mole of cyanocobalamin? What is the mass of 1.0107 molecules of cyanocobalamin? What is the mass (in grams) of one molecule of cyanocobalamin?arrow_forwardCalculate the mass in grams of each of the following samples. l type="a"> 0.994 mole of benzene, C6H6 i>4.21 moles of calcium hydride 1.79104 moles of hydrogen peroxide, H2O2 i>1.22 rnmol of glucose, C6H12O6(1mmol=11000mole) i>10.6 moles of tin 0.000301 moIe of strontium fluoridearrow_forward
- hat is the difference between the empirical and molecular for mulas of a compound? Can they ever be the same? Explain.arrow_forwardA binary compound of magnesium and nitrogen is analyzed, and 1.2791 g of the compound ¡s found to contain 0.9240 g of magnesium. When a second sample of this compound is treated with water and heated, the nitrogen is driven off as ammonia, leaving a compound that contains 60.31% magnesium and 39.69% oxygen by mass. Calculate the empirical formulas of the two magnesium compounds.arrow_forwardCalculate the percent by mass of the element listed first in the formulas for each of the following compounds. methane, CH4 sodium nitrate, NaNO3 carbon monoxide, CO nitrogen dioxide, NO2 1-octanol, C8H18O calcium phosphate, Ca3( PO4)2 3-phenyiphenol, C12H10O aluminum acetate, Al(C2H3O2)3arrow_forward
- 3.79 Consider two samples. Sample A contains 2 moles of N2 and 1 mole of O2, and Sample B contains 1 mole of N2O5. (a) Which sample contains more molecules? (b) Which sample contains more oxygen atoms? (c) Which sample contains more nitrogen atoms?arrow_forwardUsing the average atomic masses given inside the front cover of the text, calculate the mass in grams of each of the following samples. l type='a'> 5.0 moles of nitric acid 0.000305 mole of mercury 2.31105mole of potassium chromate 10.5 moles of’ aluminum chloride 4.9104moles of sulfur hexafluoride 1 25 moles of ammonia 0.01205 mole of sodium peroxidearrow_forwardYou have a 20.0-g sample of silver metal. You are given 10.0 g of another metal and told that tills sample contains twice the number of atoms as the sample of silver metal. Is this possible?arrow_forward
- erhaps the most important concept in introductory chemistry concerns what a mole of a substance represents. The mole concept will come up again and again in later chapters in this book. What does one mole of a substance represent on a microscopic, atomic basis? What does one mole of a substance represent on a macroscopic, mass basis? Why have chemists defined the mole in this manner?arrow_forwardUsing the average atomic masses given inside the front cover of this hook, calculate the mass in grains of each of the following samples. l type='a'> 0.341 mole of potassium nitride 2.62 mmol of neon (1 mmol = 1/1000 mol) 0.00449 mole of manganese(II) oxide 7.18105moles of silicon dioxide 0.00021 mole of iron(III) phosphatearrow_forwardCalculate the number of atoms of each element present in each of the following samples. l type='a'> 4.2 1 g of water 6.81 g of carbon dioxide 0.000221 g of benzene, C6H6 i>2.26 moles of C12H22O11arrow_forward
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