Download High Four Chemistry Round 2 Category C: Grades 9 – 10 Monday

Round 2
Monday, October 12, 2015
High Four Chemistry
Category C: Grades 9 – 10
The use of calculator is required.
Answer #1:
Solution:
80
1.85g
mass of sulfuric acid = 45.05 mL �
mass of sulfuric acid = 80 grams
mL
solution� �
96.0 g acid
100 g solution
�
Answer#2:
Explanation:
Evaporation / Vaporization
Evaporation / vaporization is the phase change where liquid vaporizes to
gas. The NaCl aqueous solution is a mixture of salt (solid) and water (liquid).
When heated, water evaporates and salt remains.
Answer#3:
C
Answer#4:
Solution:
27.26°C
q = mcp ∆t
945 J
q
=
= 3.76℃
∆t =
mcp (60g) �4.184 J ∙ ℃�
∆t = t final − t initial
t final = 3.76℃ + 23.5℃
t final = 27.26℃
Answer#5:
Solid
Answer#6:
Bromine trifluoride
g
Round 2
Monday, October 12, 2015
High Four Chemistry
Category C: Grades 9 – 10
The use of calculator is required.
Answer#7:
Explanation:
Sulfur
Balance the nuclear reaction. The sum of atomic numbers on the left side is
17. To obtain an equal sum of 17 on the right side, we compute: 17 – 1 =
16. The atomic number of the unknown isotope is 16.
Based on the periodic table, the atomic number 16 identifies the element
as Sulfur.
Answer#8:
Solution:
47
MWCaSO4 = 136.15 g/mol
Basis: 1 mol of CaSO4
16gO
4molsO
��
�
1molCaSO4 1molO
= 64 grams of Oxygen
grams of O = 1 mol CaSO4 �
% by mass of Oxygen =
Answer#9:
Potassium
Answer#10:
Solution:
2
grams of CH4 = 3.01x1023 H atoms �
64 grams O
x 100 = 47.01% ≈ 47%
136.15 grams CaSO4
1 CH4 mol 16g
1 H mol
CH4 �
�
�
��
4 H mol
mol
6.02283 x 1023 atoms
grams of CH4 = 2.01 grams ≈ 2 grams
Round 2
Monday, October 12, 2015
High Four Chemistry
Category C: Grades 9 – 10
The use of calculator is required.
Answer#11:
Solution:
Nitric Acid
The balanced chemical equation is as follows:
HNO3 + NaOH → NaNO3 + H2O
0.85 mol NaOH �
1 mol HNO3
� = 0.85 mol HNO3
1 mol NaOH
The theoretical number of moles of HNO3 required for this reaction is only
0.85 mol. There is an excess of 0.05 moles (the given is 0.90 mol).
Therefore, nitric acid is the excess reactant.
Answer#12:
17
Answer#13:
Solution:
9.19
Manipulate the formula ρ =
V=
Answer#14:
Solution:
0.125 kg x �
101.11
g
1000 g
1 kg
�13.6 ml�
Atomic weights of K:
N:
O:
�
m
V
to arrive at V =
m
ρ
= 9.19 mL
39.10
14.01
16.00
Molecular weight of KNO3 = 39.10 + 14.01 + 3(16.00) = 101.11 g/mol
Answer#15:
Explanation:
Beta Emission
In the given reaction, I-131 gives off a beta particle (the electron) and an
isotope.
Round 2
Monday, October 12, 2015
High Four Chemistry
Category C: Grades 9 – 10
The use of calculator is required.
Answer#16:
Solution:
4.34
mol
1L
� �1.03
� = 0.0743 mol
L
1000mL
58.44 g
grams of NaCl = 0.0743 mol �
� = 4.34 g NaCl
mol
mol of NaCl = 72.1mL �
Answer#17:
Solution:
0.90
1000g
� = 2,250 grams
1kg
100cm
length, l = 2.0 m �
� = 200 cm
1m
10cm
width, w = 0.5 dm �
� = 5 cm
1dm
thickness, t = 2.5 cm
mass, m = 2.25 kg �
Volume, V = l x w x t = (200cm)(5cm)(2.5cm) = 2,500 cm3
= 2,500 mL
Answer#18:
Explanation:
Density, ρ =
202
g
2,250 g
= 0.90
mL
2,500 mL
chemical
symbol
mass no.
(p+ + n0)
286
84
atomic no.
(p+)
X
n0 = mass no. – atomic no.
n0 = 286 – 84
n0 = 202
Round 2
Monday, October 12, 2015
High Four Chemistry
Category C: Grades 9 – 10
The use of calculator is required.
Answer #19:
Explanation:
Homogeneous
Homogeneous mixtures are mixtures that look the same throughout the
sample, even with the use of a microscope.
Answer #20:
Explanation:
2
The balanced chemical equation for the reaction is:
Mg(HCO3)2 + 2HBr
MgBr2
+
2H2O
+
2CO2
Based on the equation above, two (2) moles of carbon dioxide will be
produced.