HighFourChemistry
Category C: Grades 9 – 10
Round 9
Tuesday, May 10, 2016
The use of calculator is required.
Answer #1:
Solution:
2.60 kg
Molecular weight of FeCO3 = [55.85 + 12.01 + (16x3)] g/mol = 115.86 g/mol
Using Stoichiometric analysis:
% of Fe in FeCO3 = (
)(
) = 0.4820
kg Fe = 0.4820 ( 5400 g) (
)= 2.6028 kg
2.60 kg
2.60 kg of iron can be obtained from 5400 g of FeCO3
Answer#2:
Explanation:
Hypothesis
Hypothesis is called a tentative explanation for a set of observations based
on the data gathered.
Answer#3:
Solution:
107.8722
(
)old table = (
(
)old table = (
)present table
At of Ag old table = (
)present table
)
At of Ag old table = 107.8722
The atomic mass of silver on the table is 107.8722.
HighFourChemistry
Category C: Grades 9 – 10
Round 9
Tuesday, May 10, 2016
The use of calculator is required.
Answer#4:
Solution:
91.23 g/mol
Form the balance equation:
ZrBr4 + 4Ag 
4AgBr
+ Zr
M.W of AgBr = 187.772 g/mol
M.W of Br4 = 319.616 g/mol
Using Stoichiometric analysis:
g AgBr =
(
)(
)(
) = 23.0058
let y be the atomic mass of Zr ;
Atomic mass of ZrBr4 = y + 319.616 g/mol
12.5843g ZrBr4 = 23.0058 g AgBr(
)(
(
)(
)
)
Solving for y
y= 91.2333 g/mol
91.23 g/mol
The atomic mass of Zr obtained in the experiment is 91.23 g/mol
Answer#5:
Explanation:
Mass spectrometer
Mass spectrometer is an instrument that separates particles of different
isotopic composition and measures their individual relative masses.
HighFourChemistry
Category C: Grades 9 – 10
Round 9
Tuesday, May 10, 2016
The use of calculator is required.
Answer#6:
Solution:
7
g Na2SO4 = (15.00 – 7.05)g = 7.95 g
Tabular solution:
n(X) =
X
m(X)
MW(X)
Na2SO4
H2O
7.95 g
7.05 g
142.05 g/mol 0.0560
18.02 g/mol 0.3912
( )
( )
( )
1
6.9862 = 7
The mole ratio of H2O to Na2SO4 is 7 to 1
Empirical Formula: Na2SO4· 7H2O
X=7
From the given formula, X is equal to 7, giving an empirical formula:
Na2SO4· 7H2O.
Answer#7:
Solution:
x =2; y=2; z= 7
KXCrxOx
Assume 100 g of the compound
X
m(X)
MW(X)
K
Cr
O
26.57
35.36
38.07
39.10 g/mol
52 g/mol
16 g/mol
n(X) =
( )
( )
0.68
0.68
2.3793
( )
1
1
3.5
K2Cr2O7
From the derived empirical formula, K2Cr2O7, x=2; y=2; and z= 7
2
2
7
HighFourChemistry
Category C: Grades 9 – 10
Round 9
Tuesday, May 10, 2016
The use of calculator is required.
Answer#8:
Solution:
267.86 g
Let x = g of sulfuric acid solution
From the given, we can derive that:
56.0% sulfuric acid solution = (
Therefore, (
Solving for x:
x = 267.8571 g
)
) x = 150 g H2SO4
267.86 g
267.86 g of 56.0% sulfuric acid solution is needed to provide 150g H2SO4.
Answer#9:
Solution:
1.01 g
1 cm3 of acid has a mass of 1.45 g
Since 69.8% of the total mass of the acid is pure HNO3, then the number of
grams of HNO3 in cm3 of acid is
g HNO3 in cm3 of acid = 0.698 x 1.45 g = 1.0121 g
1.01 g
The mass of pure HNO3 in a 69.8% by weight HNO3 solution is 1.01 g.
Answer#10:
Solution:
1.5 x 10-5 cm or 1.5 x 10^-5 cm
Conversion of distance from µm to cm
Distance = 0.15 µm(
) = 1.5 x 10-5 cm
The distance in centimeter is 1.5 10-5 cm
HighFourChemistry
Category C: Grades 9 – 10
Round 9
Tuesday, May 10, 2016
The use of calculator is required.
Answer#11:
Solution:
483.12 g/mol
Determine the formula of potassium hexachloroiridate (IV)
Formula: K2IrCl6
Molar mass = [(2 x 39.10) + (6 x 35.45) + 192.22] g/mol = 483.12 g/mol
The molar mass of hexachloroiridate (IV), K2IrCl6, 483.12 g/mol
Answer#12:
Solution:
4.52 x 1023 molecules or 4.52 x 10^23 molecules
Molar mass of P4 = (4 x 30.974) = 123.896 g/mol
mole P4 = (
) = 0.7499 mol
) = 4.5159 x 1023
No. of molecules of P4 = 0.7499 mol (
No. of molecules of P4 4.52 x 1023
There are 4.52 x 1023 molecules of P4 in 92.91 g of phosphorus.
Answer#13:
Solution:
9.38 m3
Using dimensional analysis, we can formulate the formula:
m3 ocean water = (
(
3
m ocean water = (
m3 ocean water
)
(
)
)
) = 9.3846 m3
9.38 m3
9.38 m3 of ocean water must be processed to produce 0.61 kg of
bromine.
HighFourChemistry
Category C: Grades 9 – 10
Round 9
Tuesday, May 10, 2016
The use of calculator is required.
Answer#14:
Solution:
5.38 x 10-23 kg or 5.38 x 10^-23 kg
MW of C1200H2000O1000 = (12 x 1200) + 2000 + (16 x 1000) =32 400 g/mol
)(
kg C1200H2000O1000 = 1 molecule(
= 5.3803 x 10-23 kg
)(
)
5.38 x 10-23 kg
One molecule of C1200H2000O1000 has a mass of 5.38 x 10-23 kg.
Answer#15:
Solution:
0.86 mol
MW of Cd(NO3)2 ·4H2O = 308.411 g/mol
Let Y represent the formula Cd(NO3)2 ·4H2O
Using Stoichiometric analysis:
mole N = 132.4 g Y (
)(
) = 0.8586 mol
0.86 mol
There is 0.86 mol of N in 132.4 g of Cd(NO3)2 ·4H2O.
Answer#16:
Solution:
1.22 x 1024 molecules or 1.22 x 10^24 molecules
MW of Cd(NO3)2 ·5H2O = 326.411 g/mol
Let Y represent the formula Cd(NO3)2 ·5H2O
Using Stoichiometric analysis:
Molecules of H2O = 132.4 g Y (
)(
Molecules of H2O = 1.2213 x 1024 molecules
)(
1.22 x 1024 molecules
There are 1.22 x 1024 molecules of H2O in 132.4 g of Cd(NO3)2 ·5H2O.
)
HighFourChemistry
Category C: Grades 9 – 10
Round 9
Tuesday, May 10, 2016
The use of calculator is required.
Answer#17:
93.12%
Solution:
Fraction of Ag in AgCl = (
)=
= 0.7527
Mass of Ag in 7.20 g AgCl = (0.7527)(7.20 g) = 5.4194 g
Therefore the 5.82 g coin contains 5.4194 g of Ag
% of Ag in coin =
= 93.1168%
93.12 %
The coin contains 93.12% of silver in the coin
Answer#18:
Solution:
90.29%
Forming the balance equation:
Na2CO3 + CaCl2
 CaCO3
mol CaCO3 = 1.0262 g (
) = 0.010262 mol
+ 2NaCl
From the coefficients of the balance equation, mol Na2CO3 = mol CaCO3
Mass of pure Na2CO3 = 0.010262 mol(
% purity =
% purity
) = 1.087772 g
x 100% = 90.2865%
90.29%
The percentage purity of Na2CO3 in the sample is 90.29%.
HighFourChemistry
Category C: Grades 9 – 10
Round 9
Tuesday, May 10, 2016
The use of calculator is required.
Answer#19:
Solution:
8 794.5 m
Pressure of Hg = pressure of air
Formula:
Height of Hg x density of Hg x g = Height of air x density of air x g
Gravity, g, cancels out in equation (same gravity)
(0.78 m)(13 530 kg/m3) = Height of air (1.2 kg/m3)
Height of air = 8 794.5 m
The height of air needed to cause a barometer to read 78 cm of mercury
is 8 794.5 m
Answer#20:
Solution:
0.12 kg/m3
The density of a gas varies inversely with the volume, hence the formula
= is used.
d2 =
Resultant density = (0.178 6 kg/m3) (
Resultant density
) = 0.1191 kg/m3
0.12 kg/m3
The resultant density is equal to 0.12 kg/m3 as it expanded to 1.5 times its
initial volume.