Download HighFour Chemistry Round 1 Category C: Grades 9 – 10 Thursday

HighFour Chemistry
Category C: Grades 9 – 10
Round 1
Thursday, September 17, 2015
The use of calculator is required.
tin, Sn
Elements in the periodic table are
grouped according to their physical
and chemical properties of elements.
There are periodic variations in the
elements’ physical and chemical
behaviors. The periodic trend for
atomic radius is clear: it increases
from top to bottom, and decreases
from left to right.
Increasing atomic radius
Increasing atomic radius
Answer #1:
Explanation:
In the excerpt, the element with the largest atomic radius is tin (Sn).
Answer #2:
Solution:
6.43 atm
The problem is an application of the ideal gas law,
. Use the
universal gas constant, R =
(provided in Table 2 of the
Chemistry Data Booklet)
(
)(
)(
)
The pressure of the gas in the container is 6.43 atm.
Answer #3:
Explanation:
phosphorus, P
This problem requires a little bit more of analysis. Neutral atoms should
have the same number of protons and electrons. Species W, X, and Z all
have unequal numbers of protons and electrons. Specie Y is therefore
the neutral atom.
To determine the element, we know that the no. of protons is
equivalent to the atomic number because the number of protons
remains unchanged during reactions. Referring to Table 24 (Periodic
Table) of the Chemistry Data Booklet, the element with an atomic
number of 15 is phosphorus, P.
HighFour Chemistry
Category C: Grades 9 – 10
Round 1
Thursday, September 17, 2015
The use of calculator is required.
Answer #4:
Solution:
4825.765 grams Au
This is a direct application of the formula
Au (from the Periodic Table) is 196.97 g/mol.
(
Answer #5:
Solution:
. The atomic weight of
)(
)
521.7 nm
This problem is a direct application of the formula
. The velocity
or speed of light is provided in Table 2 of the Chemistry Data Booklet:
.
(
)
The green light has a wavelength of 521.7 nm.
Answer #6:
Solution:
KCN
To determine the empirical formula of this compound, convert the mass
of the elements using the atomic weights: K – 39.10; C – 12.01; N –
14.01. After calculating the no. of moles for each element, we find the
mole ratios by dividing with the least number of moles.
Element
Mass, g
Atomic Wt.,
g/mol
Mole, mol
Potassium, K
60.1
39.10
1.54
Carbon, C
18.4
12.01
1.53
Nitrogen, N
21.5
14.01
1.53
Mole Ratio
The ratio of atoms of K:C:N is therefore 1:1:1. Therefore, the empirical
formula of the compound is KCN.
HighFour Chemistry
Category C: Grades 9 – 10
Round 1
Thursday, September 17, 2015
The use of calculator is required.
Answer #7:
Explanation:
titanium (IV) chloride
Ionic compounds are made up of cations and anions. In this case, metal
cations take their names from the elements (titanium). Since titanium is
a transition metal, it occurs as Ti+2 and Ti+4. In the Stock system, Roman
numerals are used to designate different cations. The nonmetallic anions
is named by taking the first part of the element name (chlorine) and
adding “-ide”. The compound is titanium (IV) chloride.
Answer #8:
Explanation:
triple bond
A triple bond is formed when 2 atoms share
three pairs of electrons between them. The
Lewis structure for acetylene is:
H:C C:H
8e-
8eor
H−C≡C−H
Answer #9:
Solution:
58.33 mL
The problem is an application of dilution of solutions; the formula is
(VC)before = (VC)after
(
)(
)
(
)(
)
To prepare 350 mL of 1.25 M H2SO4 solution from the stock solution,
about 58.33 mL of 7.50 M H2SO4 solution.
Answer #10:
Solution:
5090.91 light year
The Avogadro’s number is a constant that can be found in Table 2 of the
Chemistry Data Booklet:
. If the thickness of a single
-3
piece of paper is 8.00 x 10 cm, and the number of pages in the book is
6.02x1023; the thickness of the book is:
(
(
)
)(
)
HighFour Chemistry
Category C: Grades 9 – 10
Round 1
Thursday, September 17, 2015
The use of calculator is required.
Answer #11:
Explanation:
7 valence eValence electrons are the outermost electrons in an atom; and these
electrons are the ones involved in chemical bonding. The valence
electron of an atom can be determined thru its electron configuration.
Bromine (Br) has an atomic number of 35, its electric configuration is:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5.
The outermost shell, n = 4, has 7 electrons. Therefore, the number of
valence electrons of bromine (Br) is 7.
Answer #12:
Solution:
92.51% 73Li
The algebraic expression for determining the % abundance of 73Li is:
(ave. atomic mass) = (%abundance x atomic mass, ) +
(%abundance x atomic mass, )
(
)
The fraction of abundance of 73Li is 0.9251. Therefore, the percentage
abundance of 73Li (0.9251 x 100%) is 92.51%.
Answer #13:
Solution:
27.2 grams NaNO3
Use the molarity formula
to determine the no. of
grams of NaNO3 crystals. The molecular weight of NaNO3 (Na = 22.99; N
= 14.01; O = 16.00) is 85 g/mol.
(
)(
)(
(
)
)
To prepare a 400-mL 0.800 M NaNO3 solution, 27.2 grams of the crystals
should be dissolved in 400 mL of distilled water.
HighFour Chemistry
Category C: Grades 9 – 10
Round 1
Thursday, September 17, 2015
The use of calculator is required.
Answer #14:
Solution:
22.01 J/mol·K
With the formula provided, the entropy of this change phase is:
This means that when 1 mole of ice melts at 0°C, there is an increase in
entropy of 22.0 J/K.
Answer #15:
Solution:
24.45 L/mol
STP stands for standard temperature (25°C) and pressure (1 atm), as
provided in Table 2 of the Chemistry Data Booklet. To solve for the molar
volume, use the formula
.
)(
(
(
Answer #16:
Solution:
)
)
8 mols HNO3
Based on the partially-balanced chemical equation, to balance out the 3
moles of Cu atoms on the left side of the reaction, there should be 3
moles of Cu ( in Cu(NO3)2 ) on the right side of the reaction. Hence,
3Cu
+
?HNO3
⟶
3Cu(NO3)2
+
?NO
+
4H2O
To balance the 8 H atoms on the right side, 8 moles of HNO3 is required.
To finalize the balancing, 2 moles of NO is required to balance out the O
and N atoms. The balanced chemical reaction is therefore:
3Cu
+
8HNO3
⟶
3Cu(NO3)2
+
2NO
+
4H2O
It requires 8 moles of nitric acid (HNO3) to balance out the reaction.
HighFour Chemistry
Category C: Grades 9 – 10
Round 1
Thursday, September 17, 2015
The use of calculator is required.
Answer #17:
Explanation:
I2 (iodine gas)
Reviewing the data provided, it is clear that iodine gas (I2) has the largest
molecular weight, and consequently, has the greatest strength of van
der Waals’ forces. It would need a higher energy to break apart these
forces. This translates into a higher boiling point.
In conclusion, iodine gas (I2) has the highest boiling point. This
conclusion can be supported by a table where their actual boiling points
are provided.
Answer #18:
Solution:
Compound
MW, g/mol
Boiling Point, °C
CH4
16.05
-164
C2H6
30.08
-89
F2
39.0
-188
C3H8
44.11
-42
C4H10
58.14
-0.5
Cl2
70.90
-34
Br2
159.8
59
I2
253.8
185
40% carbon atoms
To determine the % composition of the carbon atom, it would be best to
tabulate all data:
Atomic
# of
Mass
Element
% Composition
Weight
Atoms
Equivalent
Carbon, C
12.01
6
72.06
40
Hydrogen, H
1.01
12
12.12
6.72
Oxygen, O
16.00
6
96.00
53.28
180.18
100
HighFour Chemistry
Category C: Grades 9 – 10
Round 1
Thursday, September 17, 2015
The use of calculator is required.
Answer #19:
Explanation:
zinc (Zn)
[Ar] is called the argon core and represents 1s22s22p63s23p6. For an
electron configuration of [Ar]4s23d10, the total electrons in the element
is 30. The number of electrons in a ground-state element is also equal to
the atomic number of the element. Therefore, the element is zinc (Zn).
Answer #20:
Solution:
257.96 g/mol
The molecular weight of H5P3O10 is (H=1.01; P=30.97; O=16.00) 257.96
g/mol.