Quiz 1

Dr. Zellmer
Time: 30 mins
Name
1.
Chemistry 1250
Spring Semester 2015
Quiz I
KEY
Friday
January 30, 2015
Lab TA/time
(2 pts) You have a cup of vegetable soup. Is this most likely a homogeneous mixture, heterogeneous
mixture or a pure substance? (Explain)
Mixture - 2 or more different substances
Heterogeneous mixture - mixture whose properties are NOT uniform throughout. The
mixture of vegetables can not be made uniform, no matter
how much you mix it up. It’s properties will not be uniform
throughout. You can see that the mixture doesn’t have the same
amounts and types of vegetables throughout. While not always
the case, you can often “see” the different substances in a
heterogeneous mixture. Often the components of a
heterogeneous mixture separate out over time.
A homogeneous mixture has properties that are uniform
throughout.
2.
(4 pts) Perform the following mathematical operation and report your answer to the correct
number of significant figures. Report your answer in scientific notation. Include units.
( 4 s.f.)
103.500 (4 s.f.)
(2.115 x 103 cm2) (65.5 g + 38.000 g)
))))))))))))))))))))))))))) = 3.355856201 x 104 cm2Cg/s
6.5230 s2
(5 s.f.)
= 3.355 x 104 cm2Cg/s2 (4 s.f.)
65.5
+ 38.000
---------103.5 => 103.5 (4 s.f.)
use add/subt rule followed
by mult/div rule
(mult/div - same # s.f. as
# with least # s.f.)
This number has only 4 s.f. However, do NOT round until you get the final answer. Do this
just to determine the # s.f., but use 103.500 in the rest of the calc.
3.
(2 pts) Fill in the blanks in the table below for the isotopes indicated.
Symbol
number of number of number of
atomic
mass
protons
neutrons
electrons
number
number
)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
235 U4+
92
92
143
88
92
235
)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
A
E E = atomic symbol
Z
Z = atomic number = # p (protons)
A = mass number = #p + #n (n is neutron)
#e- = #p in a neutral atom (zero charge)
Copyright R. J. Zellmer, January 29, 2015
4.
(6 pts) The amount of mercury, Pb, in the air in a room on a particular day is 1.55 x 10!4 kg/m3.
What volume of air (in ft3) contains 3.90 x 10!3 lb of lead? (1.000 lb = 453.6 g, 1 in = 2.54 cm)
(You MUST use dimensional analysis (factor unit method) to receive full credit!)
There are 3 s.f. in 3.90 x 10!3 lb and 1.55 x 10!4 kg/m3 and 4 s.f. in 453.6 g and the other
conversions are exact so the answer has 3 s.f.
5.
(6 pts) Correctly name or give the correct formula for the following ionic compounds, using the
Stock system where necessary. (Must be legible and correctly spelled.)
(a) Cu(BrO)2
chromium(II) hypobromite
Charge on the copper ion is + 2 and is given as Roman Numeral (II) in name. You should know that
BrO! is hypobromite with a (&1) charge (BrO3! is bromate and 2 fewer O atoms gives hypobromite).
Since there are 2 hypobromites this gives a total negative charge of (&2). The copper has to provide a
positive charge of (+ 2).
XCu + 3(-1) = 0;
IO!
XCu = +2
(b) calcium dihydrogen phosphate
Charge on the calcium ion, Ca2+, is (+2). You should know that H2PO4! is dihydrogen phosphate with a
(&1) charge. Phosphate is PO43! and when you add two H+ you get H2PO4!. There needs to be 2
dihygrogen phosphate, H2PO4!, units for each Ca2+ ion for the charges to cancel.
Copyright R. J. Zellmer, January 30, 2015
(3 pts) Calculate the mass (in grams) of Bi2O3 which contains 7.42 x 1020 bismuth atoms.
(Form. wt.: Bi2O3 = 465.96 amu). Avogadro's number = 6.02 x 1023 particles/mole
6.
1 mol Bi2O3 = 6.02 x 1023 Bi2O5 f.u. = 465.96 g Bi2O3
1 Bi2O3 f.u.
465.96 g Bi2O3
? g Bi2O3 = 7.42 x 10 Bi atoms x ---------------- x ---------------------------- = 0.28716 g Bi2O3
2 Bi atoms
6.02 x 1023 Bi2O3 f.u.
= 0.287 g Bi2O3
(3 s.f.)
20
or
1 Bi2O3 f.u.
1 mol Bi2O3 f.u.
465.96 g Bi2O3
? g Bi2O3 = 7.42 x 1020 Bi atoms x ---------------- x ---------------------------- x -------------------1 mol Bi2O3
2 Bi atoms
6.02 x 1023 Bi2O3 f.u.
or
465.96 g Bi2O3
1 mol Bi atoms
1 mol Bi2O3
? g Bi2O3 = 7.42 x 1020 Bi atoms x -------------------------- x ------------------- x ------------------6.02 x 1023 Bi atoms
2 mol Bi atoms
1 mol Bi2O3
7.
(6 pts) Balance the following equation. (Must show all work. This means to show your
steps and show the atoms are balanced. Show coeff. of 1.)
2
Al(OH)3 +
3
H2SO4
v
1
Al2(SO4)3 +
6
H2O
A) Balance Al (appears in greatest amount in any cmpd., other than O, H and SO4)
2 Al(OH)3
+
H2SO4
6
Al2(SO4)3 +
H2O
B) Balance SO42- ion as a unit (appears as a SO42- unit on both sides)
2 Al(OH)3
+
3 H2SO4
6
Al2(SO4)3 +
H2O
C) Balance H
(appears in only 1 product - easier to bal. ; 12 H on left, need 12 on right)
2 Al(OH)3
3 H2SO4
+
6
Al2(SO4)3 +
6 H2O
O atoms are also now balanced (18 O atoms on left and 18 O atoms on right). You could have balanced
O atoms instead of H. Then you would ignore the O atoms in SO42- units since those O atoms already
balanced. 6-O in 2 Al(OH)3 6 need 6 H2O
Copyright R. J. Zellmer, January 30, 2015
8.
(6 pts) C7H16O3, undergoes complete combustion. Write the reaction and balance the equation.
(Must show all work. This means to show your steps and show the atoms are balanced)
2 C7H16O3 +
19 O2 v
14 CO2 +
16 H2O
A) Balance C (Appears in greatest amount in any cmpd. other than H and O for combustion reactions always balance C first)
C7H16O3 +
O2
6
7 CO2 +
H2O
(Note: # CO2 = # C’s in the reactant molecule for
combustion of a hydrocarbon)
B) Balance H (leave oxygen until end)
(Note: # H2O = ½ # H’s in hydrocarbon)
C7H16O3 +
C) Balance O
C7H16O3 +
O2
6
7 CO2 +
8 H2O
22 O atoms in products and 3 O atoms in the C7H16O3 - need
19 O atoms from O2 - use a fraction
6
19/2 O2
7 CO2 +
8 H2O
taking ½ of 19 O=O molecules gives 19 O atoms
In the fraction the numerator equals the number of atoms you need and the denominator
equals the subscript for that element in the compound. If this had been O3 you would
have used 19/3.
D) Multiply through by 2 (want whole number coefficients)
2 C7H16O3 +
19 O2
6
Copyright R. J. Zellmer, January 30, 2015
14 CO2 +
16 H2O
9.
(5 pts) Calculate the mass percent composition of nitrogen in Mg(NO2)2.
(At. wts: Mg = 24.31, N = 14.01, O = 16.00) (Must show all work.)
Mass % of =
element
mass element
-------------------- x 100%
total mass
mass element
mass % = ------------------- x 100%
MW (or FW)
From a molecular or :
empirical formula
Mg : 1 x 24.31 amu = 24.31 amu
N : 2 x 14.01 amu
= 28.02 amu
O : 4 x 16.00 amu
= 64.00 amu
--------------= 116.33 amu (5 s.f.)
FW for Mg(NO2)2
24.31 amu
% Mg = ----------------- x 100 % = 20.90 % Mg (4 s.f.)
116.33 amu
28.02 amu
% N = ----------------- x 100 % = 24.09 % N (4 s.f.)
116.33 amu
64.00 amu
% O = ----------------- x 100 % = 55.02 % O (4 s.f.)
116.33 amu
Copyright R. J. Zellmer, January 30, 2015

Download Report