CHAPTER 2 REGULAR HOMEWORK
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Question 1
In one on the compounds that contains only oxygen and nitrogen, 3.50 g of nitrogen combines with 2.00 g of oxygen. In another, 0.875 g of nitrogen combines with 1.00 g of oxygen. Show that these compounds obey the law of multiple proportions. Select one: 
Explanation
The law of multiple proportions states that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers. For example, there are five distinct oxides of nitrogen, and the weights of oxygen in combination with 14 grams of nitrogen are, in increasing order, 8, 16, 24, 32, and 40 grams, or in a ratio of 1, 2, 3, 4, 5. In the first compound the ratio of oxygen to nitrogen is 1: 1.75 In the second compound the ratio of oxygen to nitrogen is 1: .875 The difference between those two compounds is in the amount of nitrogen in while the first has double the amount of the second making it a 2:1. (which is exactly what the answer choice b is implying). To look at this in depth click this!
Question 2
Chromium forms two bromides, one that is 17.82% chromium by mass and another that is 24.55% chromium. You will show that these compounds obey the law of multiple proportions. In the first sample, what is the mass of chromium per 1 gram of bromide? (Hint, in a 100 g sample how many grams are chromium and how many grams are bromide? The chromium to bromide ratio gives the mass of chromium per 1 gram of bromide.)
Explanation
In a 100g sample… The first bromide is 17.82 g Chromium and is 82.18 g Bromine Ratio: 1: 4.6116 in accordance to Chromium (17.82/17.82 && 82.18/17.82) Ratio: .216:1in accordance to Bromine(17.82/82.18 && 82.18/82.18)
Question 3
In the second sample, what is the mass of chromium per 1 gram of bromide?
Explanation
The second bromide us 24.55 g Chromium and is 75.45 g Bromine Ratio: 1: 3.07 in accordance to Chromium Ratio: .325:1
Question 4
What is the ratio of mass of chromium in the two samples per gram of bromide? (Use the ratio of the larger mass to the smaller mass.) Note, you must enter a number for you answer. 1:1 or 1/1 = 1. So enter 1. 5:4 or 5/4 = 1.2, 4:3 or 4/3 = 1.33, 3:2 or 3/2 = 1.5, etc.
Explanation
Note that this ratio is equivalent to the ratio of small whole numbers. (Divide your answer by 1 and multiple numerator and denominator by 2.) .325 : .216.325/.216 = 1.50
Question 5
Given the symbol 127I-. What is the atomic number?
Explanation
The atomic number (proton number) of an element will never change, because if it did than the element itself would change. So find iodine on a periodic table and find its atomic number.
Question 6
Given the symbol 127I-. What is the mass number?
Explanation
Mass number is also not dependent on the number of electrons as they are simply negligible in comparison to neutrons and protons, so the mass number of this element will be equivalent to its atomic number (proton count) and its number of neutrons, which just so happens to be shown here on the top left corner (127).
Question 7
Given the symbol 127I-. What is the charge?
Explanation
There is an extra electron in this Iodine and it has a charge of -1 which can be seen on the top right corner of elemental I.
Question 8
Given the symbol 127I-. What is the number of protons?
Explanation
I hope you’ve finally figured it out by now…it’s the same exact thing as it’s atomic number, which you already found.. surprise surprise it’s 53.
Question 9
Given the symbol 127I-. What is the number of electrons?
Explanation
If this element was in its ground state than it would have the same number of electrons as it does protons, but since it’s in an excited state as an ion and has a charge of -1, it has an additional electron making it 53 + 1, which is 54. Ya got the gist!
Question 10
Given the symbol 127I-. What is the number of neutrons?
Explanation
Take the mass number and subtract it from the atomic number aka take 127 – 53 = 74.
Question 11
Given a mass number of 17, a charge of 2-, and 10 electrons. What is the atomic number?
Explanation
A charge of 2- means there was originally only 8 electrons (8 + 2 =10) and usually in the ground state the number of electrons and protons are the same, so that means there was 8 protons, which makes 8 the atomic number as the two are synonymous.
Question 12
Given a mass number of 17, a charge of 2-, and 10 electrons. What is the number of protons?
Explanation
Ya’ll at this point if you haven’t figured out that the atomic number and the number of protons is the same I DONT KNOW WHAT I CAN DO!
Question 13
Given a mass number of 17, a charge of 2-, and 10 electrons. What is the number of neutrons?
Explanation
With a mass number of 17 and an atomic number/# of protons being 8: 17-8=9 giving us 9 neutrons.
Question 14
Given a mass number of 17, a charge of 2-, and 10 electrons. What is the element? Select one:
Explanation
Protons/Atomic # will give you the element and in this case it’s oxygen.
Question 15
Chlorine has two stable isotopes of mass 34.97 u and 36.97 u. Suppose the percent abundance of the lighter isotope were 67.22%. What would be the average atomic mass of Cl? (The actual abundance is 75.77%.) USE 4 SIG. FIG.
Explanation
34.97 ( .6722) + 36.97 (.3278) = 35.62
Question 16
The mass spectrum of a hypothetical element shows that 75.58% of the atoms have a mass of 27.977 u, 1.522% have a mass of 28.976 u, and the remaining have a mass of 29.974 u. Calculate the average atomic mass of this element (in units of u). (Use 4 Sig. Fig. Don’t round until end.)
Explanation
4 Sig. Fig. (Don’t round until end.) 27.977(.7558) + 28.976(.01522) + 29.974( .22898)= 28.45
Question 17
Hypothetical element X has 3 stable isotopes. The relative peak intensities are 20.00 for 180.0 u, 100.0 for 184.0 u, and 60.00 for 186.0 u. What is the average atomic mass for X? USE THE CORRECT NUMBER OF SIG. FIG.
Explanation
(180*20 + 100*184 + 60*186) / (20 + 100 + 60) = 184.2
Question 18
Bromine occurs naturally as two isotopes, one with a mass of 78.918 u and the other with a mass of 80.916 u. Suppose the average atomic mass from the periodic table were 80.77 u. What would be the percent abundance of the lighter isotope? (Use, 4 SIG. FIG.) FYI (not useful for solving the problem) the actual mass and % abundance are 79.904 u and 50.65%, respectively.
Explanation
solve using a system of equations 80.77 = 78.918x + 80.916y 1 = x + y 1-y =x 80.77 = 78.918(1-y) + 80.916y y= .92693 x= .07307 y=.92693
Question 19
Calculate the molecular mass (in u) of phosphoric acid, H3PO4 to 4 significant figures.
Explanation
Get the molar masses of each individual atom and multiply it by the subscript and add them all up. H-1.008 * 3 = 3.024 P- 30.97 O-16*4 = 64 97.9
Question 20
Calculate the molecular mass (in u) of aniline, C6H7N to 4 significant figures.
Explanation
Get the molar masses of each individual atom and multiply it by the subscript and add them all up. C-12.0107 * 6 = 72.06 H-1.008 * 7 = 7.056 N-14* 1 = 14 93.11/93.12
Question 21
What charge is expected for the following elements in ionic compounds–Be, Mg, Ca, Sr, Ba, Ra, Zn, and Cd? For positive charges write: 1 or +1, 2 or +2, etc. For negative charges write: -1, -2, etc. DO NOT WRITE: 1-, etc.
Explanation
+2 look at the charges periodic table above or linked here
Question 22
What charge is expected for the following elements in ionic compounds–O, S, Se, Te, and Po?
Explanation
-2 look at the charges periodic table above or linked here
Question 23
What charge is expected for the following elements in ionic compounds–Li, Na, K, Rb, Cs, and Fr?
Explanation
+1 look at the charges periodic table above or linked here
Question 24
What is the expected empirical formula for the ionic compound made from Li and I? Select one:
Explanation
Lithium has a 1+ charge Iodine has a 1- charge 1:1
Question 25
What is the expected empirical formula for the ionic compound made from Cs and O? Select one:
Explanation
Oxygen has a 2- change Cesium has a 1+ charge O=1 Cs=2
Question 26
What is the empirical formula for the compound made from Y3+ and Cl1-? Select one:
Explanation
Always balance the charges to make it equal 0. Y = 3+ Cl=1- 3+ + 3(1-) = 0 1Y 3Cl
Question 27
Calculate the number of moles in 12.7 g of Se.
Explanation
3 Sig. Fig. 12.7 / 78.96 = .16
Question 28
Calculate the number of moles in 1.33×1022 NO2 molecules.
Explanation
3 Sig. Fig. 1.33 * 10^22 / (6.022*10^23) = .0220
Question 29
Calculate the mass (in g) of 1.41 moles of H2O.
Explanation
g/ (18 MM ) = 1.41mol g = 25.38
Question 30
Calculate the number of moles in 13.7 g SO3.
Explanation
80.066 MM of SO# 13.7/80.066 = .171 mol
Question 31
Calculate the number of oxygen atoms in 77.8 g SO3.
Explanation
80.066= MM of SO3
77.8 g / 80.066 g = .972 mol
.972 mol * 6.022 * 10^23 * 3 (multiply the number by 3 because there was 3 oxygen atoms on every mol of SO3) ==1.75e24