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Quiz von kiaraisastudent, aktualisiert more than 1 year ago
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Frage 1
Frage
Calculate the mass of 1.000 mole of CaCl(2).
Antworten
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110.986 g/mol
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118.56 g/mol
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342.15 g/mol
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106.323 g/mol
Frage 2
Frage
Calculate grams in 3.0000 moles of CO(2).
Antworten
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132.03 g
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102.01 g
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124.09 g
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137.15 g
Frage 3
Frage
Calculate number of moles in 32.0 g of CH(4).
Antworten
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1.87 mol
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1.99 mol
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2.11 mol
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2.03 mol
Frage 4
Frage
Determine mass in grams of 40.0 moles of Na(2)CO(3).
Antworten
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4,240 g
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5,565 g
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4,145 g
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4,056 g
Frage 5
Frage
Calculate the percent composition of KNO(3).
Antworten
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Molar mass = 151.1 g/mol Potassium: (39.10 / 101.1) x 100 = 38.67% Nitrogen: (14.01 / 101.1) x 100 = 13.86% Oxygen: (48.00 / 101.1) x 100 = 12.58%
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Molar mass = 164.1 g/mol Potassium: (39.10 / 101.1) x 100 = 38.67% Nitrogen: (14.01 / 101.1) x 100 = 18.96% Oxygen: (48.00 / 101.1) x 100 = 47.48%
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Molar mass = 101.1 g/mol Potassium: (39.10 / 101.1) x 100 = 38.67% Nitrogen: (14.01 / 101.1) x 100 = 13.86% Oxygen: (48.00 / 101.1) x 100 = 47.48%
Frage 6
Frage
Calculate percent composition for H(2)SO(4).
Antworten
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Molar mass = 98.07 g/mol Hydrogen: (2.016 / 98.07) x 100 = 2.06% Sulfur: (32.06 / 98.07) x 100 = 32.69% Oxygen: (64.00 / 98.07) x 100 = 65.26%
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Molar mass = 98.07 g/mol Hydrogen: (2.016 / 98.07) x 100 = 3.09% Sulfur: (32.06 / 98.07) x 100 = 54.12% Oxygen: (64.00 / 98.07) x 100 = 65.39%
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Molar mass = 98.07 g/mol Hydrogen: (2.016 / 98.07) x 100 = 9.05% Sulfur: (32.06 / 98.07) x 100 = 44.75% Oxygen: (64.00 / 98.07) x 100 = 14.96%
Frage 7
Frage
An oxide of chromium is found to have the following percent composition: 68.4% Cr and and 31.6% O. What is the compound's empirical formula?
Antworten
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CrO(4)
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Cr(6)O(2)
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Cr(2)O(3)
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Cr(5)O
Frage 8
Frage
A sample with a molar mass of 34.00 g/mol is found to consist of 0.44g H and 6.92g O. Find its molecular formula.
Antworten
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HO
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H(2)O
Frage 9
Frage
The percent composition of a compound was found to be 63.5% silver, 8.2% nitrogen, and 28.3% oxygen. Determine the compound's empirical formula.
Antworten
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Ag(3)NO
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Ag(2)NO(3)
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Ag(4)NO
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AgNO(3)
Frage 10
Frage
If 4.04g of N combine with 11.46g O to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound?
Antworten
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N(3)O(6)
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N(2)O(5)
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N(1)O(8)
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N(2)O(3)
Frage 11
Frage
[blank_start]Percent composition[blank_end] is the percent by mass of each element in a compound.
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Percent composition
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Molar mass
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Molecular formula
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Empirical formula
Frage 12
Frage
[blank_start]Avogadro's number[blank_end] is the number 6.02 * 10^23, which is the number of representative particles in a mole.
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Percent Composition
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Avogadro's number
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Molecular Formula
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Molar Mass
Frage 13
Frage
[blank_start]Molar mass[blank_end] is the mass in grams of one mole of any pure substance.
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Molar mass
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Molecular Formula
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Avogadro's Number
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Percent Composition
Frage 14
Frage
[blank_start]mole[blank_end]: the SI base unit used to measure the amount of a substance. It is the number of representative particles, carbon atoms, in exactly 12 g of pure carbon-12.
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molar mass
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mole
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hydrate
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empirical formula
Frage 15
Frage
[blank_start]A hydrate[blank_end] is a compound that has a specific number of water molecules bound to its atoms.
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A hydrate
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Molar mass
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A mole
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Empirical formula
Frage 16
Frage
[blank_start]Molecular formula[blank_end] is a formula that specifies the actual number of atoms of each element in one molecule or formula unit of the substance.
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Empirical formula
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Molecular formula
Frage 17
Frage
[blank_start]Empirical formula[blank_end] is a formula that shows the smallest whole number mole ratio of the elements of a compound and may or may not be the same as the [blank_start]molecular formula[blank_end].
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Empirical formula
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Molecular formula
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empirical formula
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molecular formula
Frage 18
Frage
Calculate percent composition of C(2)H(5)OH.
Antworten
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Molar mass = 46.07 g/mol Carbon: (24.022 / 46.07) x 100 = 12.64% Hydrogen: (6.048 / 46.07) x 100 = 13.13% Oxygen: (16.00 / 46.07) x 100 = 94.72%
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Molar mass = 55.07 g/mol Carbon: (24.022 / 46.07) x 100 = 22.55% Hydrogen: (6.048 / 46.07) x 100 = 14.13% Oxygen: (16.00 / 46.07) x 100 = 34.73%
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Molar mass = 86.07 g/mol Carbon: (24.022 / 46.07) x 100 = 52.14% Hydrogen: (6.048 / 46.07) x 100 = 13.13% Oxygen: (16.00 / 46.07) x 100 = 34.73%
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