Chemistry tutorials

General Chemistry Software Download

1) Periodic Table

2) Molecular Calculator

3) Balance

4) Thermochemistry

5) Solubility Product

6) Weak acid/base, buffers

7) Electrochemistry

8) Spectroscopy

9) Curve Fit & Function Plot

10) Molecular Structure (Molecular 3-D Viewer)

11) Gas Equations

12) Conversions

13) Ternary Plot

Periodic Table

1.1 Periodic Table - Symbols:
What is the symbol for:
Potassium
Hydrogen
Sodium
Calcium
Sulfur
Oxygen
Carbon
Lithium
Manganese
Tin


Solution: Activate radio button Name and compare push button symbols by edit field names.


1.2 Periodic Table - Names
Find names for following symbols:
Br
Sn
Ba
H
Cl
C
P
Al
Si
K
Fe
Pb

Solution: Activate radio button Name and compare edit field names by push button symbols.
1.3 Periodic Table - Atomic number
Find the atomic number of:
Uranium
Lead
Phosphorus
Zinc
Tin
Iodine
Nitrogen
Oxygen
Sodium
Neon
Boron
Chromium

Solution: Activate radio button Atomic number.
1.4 Periodic Table - Acid-base properties (oxides)
Find the properties (acid-amph.-basic) for the oxides of the following elements:
Sulfur
Potassium
Chlorine
Manganese
Fluorine
Aluminum
Iodine
Sodium
Boron
Nickel
Copper
Lithium

Solution: Activate radio button Acid-base properties..
1.5 Periodic Table - Phase/state:
Where can we locate the gases in the periodic table.

Solution: Activate radio button Phase and search for "Gas" in the edit fields.
1.6 Properties - Isotopes
Find the number of stable isotopes and their abundance:
Hydrogen
Carbon
Sodium
Tin

Solution: Select push buttons H, C ... and activate radio button Stable isotopes in "Properties".
1.7 Properties - History
Oxygen: Explain the bright red and yellow-green colors of the Aurora
Hydrogen: How many stable and unstable hydrogen isotopes exist. What are their names ?
Uranium: What uranium isotope causes fission (nuclear energy).
Platinum: What will happen if platinum is inserted in an oxygen-hydrogen atmosphere ?
Nitrogen: Find atmospheric vol.% of nitrogen. Determine formula: ammonia, potassium nitrate and ammonium nitrite.
Carbon: Name three allotropic forms of carbon ?
Helium: Describe the use of liquid helium.

Solution: Select push buttons O, H ... and activate radio button History in "Properties" .
1.8 Explain the relationship/connection: electronegativity/atomic radius:

Solution: Alter between the radio buttons Elecronegativity/Atomic radius.
1.9 Decide max/min values of:
Electronegativity
Atomic radius
Melting point
Boiling point
Electrical conductivity
Thermal conductivity
Number of stable isotopes
Density

Solution: Activate Graphics in the periodic table. Use radio buttons (electronegativity...) and search for max/min values.

Back to top

Molecular Calculator

2.1 Calculate moles:
a) 12 grams of NaCl
b) 23 grams of CaO
c) 9.43 grams of CuSO₄*5H₂O

Solutions:
a) 12gNaCl
b) 23gCaO
c) 9.43gCuSO4*5H2O

2.2 Calculate grams:
a) 0.1 moles of CuCl₂
b) 0.5 moles of NH₄Cl
c) 2 moles of CaSO₄*5H₂O

Solutions:
a)1mCuCl2
b) 0.5mNH4Cl
c) 2mCaSO4*5H2O

2.3 Calculate moles of carbon (C) in:
a) 3.2 moles of CH₄
b) 0.5 moles of HCN
c) 2.17 moles of K₄Fe(CN)₆
Solutions:
a) 3.2mCH4
b) 0.5mHCN
c) 2.17mK4Fe(CN)6

2.4 Calculate grams of oxygen (O) in:
a) 3 grams of CaO
b) 21 grams of H₂O₂
c) 0.23 grams of BaSO₄

Solutions:
a) 3gCaO
b) 21gH2O2
c) 0.23gBaSO4

2.5 Calculate moles of hydrogen (H) in:
a) 1.4 grams of LiH
b) 0.3 grams of H₂O
c) 12 grams of CH₃CH₂OH

Solutions:
a) 1.4gLiH
b) 0.3gH2O
c) 12gCH3CH2OH

2.6 Calculate grams of Sodium (Na) in:
a) 0.6 moles of NaCl
b) 3.4 moles of Na₂CO₃
c) 2.3 moles of NaH₂PO₄

Solutions:
a) 0.6mNaCl
b) 3.4mNa2CO3
c) 2.3mNaH2PO4

2.7 Calculate % Calcium (Ca) in:
a) 2.1 moles of CaO
b) 0.6 moles of CaCl₂
c) 3.2 moles of Ca₃(PO₄)₂

Solutions:
a) 2.1mCaO
b) 0.6mCaCl2
c) 3.2mCa3(PO4)2

2.8 Calculate % Fluorine (F) in:
a) 1.2 grams of HF
b) 2.7 grams of MgF₂
c) 0.9 grams of AlF₃

Solutions:
a) 1.2gHF
b) 2.7gMgF2
c) 0.9gAlF3

2.9 Calculate:
a) % K, Cr and O in K₂CrO₄.
b) Grams of: Ca, C and O in 0.3 mol CaCO₃.
c) Moles of Pb and O in 12grams PbO₂.

Solutions:
a) K2CrO4.
b) 0.3mCaCO3.
c) 12gPbO2.

2.10 Calculate grams and moles (all elements) in Mg₃(PO₄)₂.

Solution:
Mg3(PO4)2

2.11 Calculate moles:
a) 2 moles of Na in NaCl
b) 0.3 moles of Cl in AlCl₃
c) 3.6 moles of P in Ca₃(PO₄)₂
Solutions:
a) 2mNa,NaCl
b) 0.3mCl,AlCl3
c) 3.6mP,Ca3(PO4)2

2.12 Calculate grams of:
a) CaO containing 2 grams of O
b) HCl containing 5 grams of Cl
c) Fe₂O₃ containing 2.4 grams of Fe

Solution:
a) 2gO,CaO
b) 5gCl,HCl
c) 2.4gFe,Fe2O3

2.13 Calculate moles of Potassium (K) in:
a) KNO₃ containing 3.2 moles of O
b) KClO₄ containing 0.2 moles of Cl
c) CaCl₂ containing 1.4 moles of Ca

Solutions:
a) 3.2mO,KNO3
b) 0.2mCl,KClO4
c) 1.4mCa,CaCl2

2.14 Calculate grams of Nitrogen(N) in:
a) NO containing 12 grams of O
b) NH₃ containing 0.5 grams of H
c) NH₄Cl containing 3.0 grams of Cl

Solutions:
a) 12gO,NO
b) 0.5gH,NH3
c) 3.0gCl,NH4Cl

2.15 Calculate moles of barium (Ba) in:
a) BaO containing 5.4 grams of O
b) BaCl₂ containing 3.7 grams of Cl
c) BaCO₃ containing 1.0 grams of C

Solutions:
a) 5.4gO,BaO
b) 3.7gCl,BaCl₂
c) 1.0gC,BaCO3 2.16 Calculate grams of Cu in:

2.16 Calculate grams of Cu in:
a) CuS containing 0.6 moles of S
b) CuCl₂ containing 6.4 moles of Cl
c) CuSO₄*5H₂O containing 7.8 moles of O
Solutions:
a) 0.6mS,CuS
b) 6.4mCl,CuCl2
c) 7.8mO,CuSO4*5H2O

2.17 Calculate % bromide (Br) in:
a) NaBr containing 2.1 moles of Na
b) CaBr₂ containing 3.0 moles of Ca
c) AlBr₃ containing 0.2 moles of Al

Solutions:
a) 2.1mNa,NaBr
b) 3.0mCa,CaBr2
c) 0.2mAl,AlBr3

2.18 Calculate % lithium (Li):
a) LiI containing 9.0 grams of I
b) Li₂O containing 2.8 grams of O
c) LiHCO₃ containing 1.0 grams of C

Solutions:
a) 9.0gI,LiI
b) 2.8gO,Li2O
c) 1.0gC,LiHCO3

2.19 Calculate
a)% Na, C and O if Na₂CO₃ contains 4 grams of O.
b)Grams of iron and oxygen in Fe₂O₃ containing 0.8 moles of Fe.
c)Moles of lead, carbon and hydrogen in Pb(C₂H₅)₄ containing 4.0 grams of hydrogen.

Solutions:
a) 4gO,Na2CO3
b) 0.8mFe,Fe2O3
c) 4gH,Pb(C2H5)4

2.20 Calculate %, grams and moles of all elements in C₄H₉OH.

Solution:
C4H9OH

2.21 Calculate max.amount of formed Ca₃(PO₄)₂ using 3 grams of Ca ,10 grams of O and 14 grams of P?

Solution:
Determine by calculating following argumented formulas: 3gCa,Ca3(PO4)2 and 10gO,Ca3(PO4)2 and 14gP,Ca3(PO4)2

2.22 How many moles of Be must be used to make 245 grams BeCl₂?

Solution:
245gBeCl2

2.23 The CuSO₄*5H₂O -compound release water when heated. How much mass will evaporate in this process?

Solution:
mass of (CuSO4*5H2O) - mass of (CuSO4) = mass of 5H2O

2.24 Determine empirical formula:
a) 92.83% lead and 7.67% oxygen.
b) 31.9% potassium, 28.9% chlorine and 39.2% oxygen.

Solution: Vary element composition of formula and investigate the element ratio. It will simplify calculation if one of the elements in the trial formulas are argumented by grams.
a) First argumented trial formula , 92.83gPb,PbO shows a plausible element mass ratio.
b) First argumented trial formula, 31.9gK,KClO indicates to little oxygen. Second argumented trial formula 31.9gK,KClO2 also indicates to little oxygen. Third argumented formula 31.9gK,KClO3 shows a correct element ratio.

2.25 0.3 grams of metallic silver was dissolved. A precipitate (0.399 grams of AgX) formed after adding a unknown substance. Determine X knowing that X=Cl- or X=Br-.

Solution: Determine by the use of argumented trial formulas (0.3g) and compare results: 1) 0.3gAg,AgBr or 2) 0.3gAg,AgCl .

2.26 A solution of dissolved Ca(OH)₂ was added CO₂. A white precipitate formed. The precipitate was filtered, dried and weighed. (2.35 grams). How many grams of Ca was involved in the precipitation process?

Solution:
Ca(OH)2+CO2>CaCO3+H2O
2.35gCaCO3

2.27 2 grams of Al was added HCl:
2Al(s) + 6HCl(aq)=2AlCl₃ + 3H₂
a) How many grams of AlCl₃ was formed in the reaction?
b) How many moles of HCl was used forming AlCl₃?


Solutions:
a) 2gAl,AlCl3
b) moles of AgCl3*3= moles of HCl

2.28 3 grams of Fe reacts with 1.29 grams of O₂ forming a pure iron compound.
a) Decide the empiric formula?


Solution:a)First argumented trial formula, 3gFe,FeO contains to small amount of oxygen. Second argumented trial formula, 3gFe,FeO2 contains to much oxygen. Third argumented trial formula, 3gFe,Fe2O2 contains to small amount of oxygen. At least, the formula 3gFe,Fe2O3 shows a correct iron/oxygen ratio.

Back to top

Balance

3.1 Balance:
a) Cu+O2>CuO
b) Fe+Cl2>FeCl3
c) C+Br2>CBr4

Solutions: Copy bal.expr. into "unbalanced equation" edit field and calculate.

3.2 Balance:
a) Zn+HCl>ZnCl2+H2
a) Al+HCl>AlCl3+H2
c) Sn+HCl>SnCl4+H2

Solutions: Copy bal.expr. into "unbalanced equation" edit field and calculate.


3.3 Alkane-oxygen combustion's generally forms CO2 and water. Balance:
a) CH4+O2>CO2+H2O
b) C2H6+O2>CO2+H2O
c) C3H8+O2>CO2+H2O

Solutions: Copy bal.expr. into "unbalanced equation" edit field and calculate.

3.4 Balance:
a) Na2CO3+HNO3>NaNO3+H2O+CO2
b) KClO3+S+H2O>Cl2+K2SO4+H2SO4
c) FeS2+O2>Fe3O4+SO2
d) Al(OH)3+H2SO4>Al2(SO4)3+H2O
e) KBr+MnO2+H2SO4>Br2+MnBr2+KHSO4+H2O
f) Na3SbS4+HCl>Sb2S5+H2S+NaCl
g) Cu+HNO3>Cu(NO3)2+NO+H2O
h) Ca3P2+H2O>Ca(OH)2+PH3
i) FeSO4+KMnO4+H2SO4>Fe2(SO4)3+MnSO4+K2SO4+H2O

Solutions: Copy bal.expr. into "unbalanced equation" edit field and calculate.

3.5 8 grams of C5H12 reacts with oxygen.
Balance:
C5H12+O2>CO2+H2O and determine:
a) How many grams of CO2 was formed in the reaction?
b) How many grams of H2O was formed in the reaction?
c) How many grams of O2 was used during the reaction?

Solutions:8gC5H12+O2>CO2+H2O Copy argumented bal.expr. into "unbalanced equation" edit field and calculate.


3.6 A piece of metallic iron (3g) was dissolved in conc.HCl (6 grams of 100% HCl). The reaction formed H₂ and FeCl₂.
Balance: Fe+HCl>FeCl2+H2 and determine
a) amount of formed FeCl₂
b) amount of formed H₂


Solutions: 3gFe+6gHCl>FeCl2+H2 Copy argumented bal.expr. into "unbalanced equation" edit field and calculate.


3.7 Balance: N₂+H₂>NH₃ and determine how many grams of nitrogen and hydrogen used in a reaction forming 15.0 grams of NH3 .

Solution:N2+H2>15gNH3 Copy argumented bal.expr. into "unbalanced equation" edit field and calculate.

3.8 Balance following heated mixture reaction: As₂S₃+NaNO₃+Na₂CO₃>Na₃AsO₄+Na₂SO₄+NaNO₂+CO₂
Determine:
a) Amount of As₂S₃, NaNO₃ and Na₂CO₃ used forming 23.0 grams of Na₃AsO₄ ?
b) How much CO₂ will be formed if the "raw material" for the reaction was 10.0 grams As₂S₃, 75.0 grams NaNO₃ and 37.0 grams of Na₂CO₃ ?

Solutions:Modify balance expr. and calculate
a) As2S3+NaNO3+Na2CO3>23gNa3AsO4+Na2SO4+NaNO2+CO2
b) 10gAs2S3+75gNaNO3+37gNa2CO3>Na3AsO4+Na2SO4+NaNO2+CO2


3.9 Balance:
Na₂CO₃+HNO₃>NaNO₃+H₂O+CO₂
Determine:
a) How much carbon dioxide will be formed using 13.4 grams of sodium in Na₂CO₃ ?
b) How much H₂O will be formed if Na₂CO₃ contains 3.4 grams of carbon and HNO₃ contains 18.8 grams of nitrogen?
c) How many moles of NaNO₃ will be formed if CO₂ contains 12.3 grams of carbon.

Solutions:Modify balance expr.:
a)13.4gNa,Na2CO3+HNO3>NaNO3+H2O+CO2
b)3.4gC,Na2CO3+18.8gN,HNO3>NaNO3+H2O+CO2
c)Na2CO3+HNO3>NaNO3+H2O+12,3gC,CO2

3.10 Metallic iron may be produced by heating a mixture of Fe2O3 and carbon.
Balance:
Fe₂O₃+C>Fe+CO₂
and determine moles of:
a) Fe₂O₃ in a reaction forming 1000 grams of Fe.
b) Fe in a reaction using 200 grams of C.
c) CO₂ in a reaction forming 12.8 moles Fe.

Solutions:Modify balance expr.
a) Fe2O3+C>1000gFe+CO2
b) Fe2O3+200gC>Fe+CO2
c) Fe2O3+C>12.8mFe+CO2


3.11 FeS₂ and O₂ reacts forming Fe₃O₄ and SO₂. Balance:
FeS₂+O₂>Fe₃O₄+SO₂
Determine the amount of formed SO₂ and Fe₃O₄ using 12.0 grams of Fe in FeS₂ and 1.5 moles of O in O₂?.

Solution:Modify balance expr.
12gFe,FeS2+1.5mO,O2>Fe3O4+SO2


3.12 KClO3 added HCl forms Cl₂
Balance:
KClO₃+HCl>H₂O+KCl+Cl₂
Determine:
a)Amount of KClO₃ used in a reaction forming 2 grams of Cl₂.
b)A compound (KClO₃) contains 4 g oxygen. How much Cl₂ will be formed adding 0.04 moles of HCl ?

Solutions:Modify balance expr.
a) KClO3+HCl>H2O+KCl+3gCl2
b) 4gO2,KClO3+0.04mHCl>H2O+KCl+Cl2


3.14 1 kg octane reacts with oxygen. Determine vol. O2 (NTP) used in this reaction.

Solution:Modify balance expr.
Step 1: 1000gC8H18 + O2 > CO2 +H2O
Step 2: Calculate O2 -vol. using V=nRT/p.
(n) moles of oxygen (calculated in Step 1):
(gas const.) R=0.082liter*atm/(Kelvin*mol)
(temp.) T=273.15Kelvin
(pressure) p=1atm.


3.15 10.99 grams of a compund containing C, H and O reacts with O₂ forming H₂O and 21.0 grams of CO₂. What's the empiric formula for the compund?


Solution:Find the answer by observing the effect of varying the values of X,Y and Z in following argumented (21g) balance expression:
CXHYOZ + O2 > 21gCO2 + H2O

Back to top

Thermochemistry

4.1 Determine H for the combustion of (C₄H₁₀) and (O₂).

Solution: Insert C4H10(g)+O2(g)>CO2(g)+H2O(l) and calculate

4.2 Determine H when 2 grams Pb (lead) reacts with oxygen forming PbO?

Solution: Insert 2gPb(s)+O2(g)>PbO(s) and calculate

4.3 Balance the equation: Fe(s)+O₂(g)>Fe₂O₃(s).
a)Determine H
Is this reaction exothermic/endothermic?. b)Determine S
c)Determine H and S if 2 grams of Fe₂O₃ is formed.

Solutions:
a) and b)Insert Fe(s)+O2(g)>Fe2O3(s) and calculate.
c)Insert the modified equation: Fe(s)+O2(g)>2gFe2O3(s) and calculate.

Back to top

Solubility Product

5.1 Decide Ksp-expr.:
a) PbF₂
b) Ag₂CrO₄

Solutions:
a) Insert eq. of dissociation: PbF2 > Pb+2 + 2F- and calculate the Ksp-expr.
b) Insert eq. of dissociation: Ag2CrO4 > 2Ag+ + CrO4-2 and calculate the Ksp-expr.

5.2 Determine Ksp Ca(OH)₂ if 0.0105 mol dissolves in 1 kg of water.

Solution:
Step 1) Insert equation of dissociation: Ca(OH)2 > Ca+2 + 2OH-
Step 2) Insert 0.0105 in the [Ca⁺²] field (Mass solvent = 1 kg)
Step 3) Calculate

5.3 How many grams of CdF₂ dissolves in 1 kg of water when [F^-] = 0.234 M ?

Solution:
Step 1) Insert equation of dissociation: CdF2 > Cd+2 + 2F-
Step 2) Insert 0.234 in the [F- ] field (Mass solvent = 1 kg)
Step 3) Calculate (CdF2)

5.4 How many grams of PbCl₂ dissolved in 1 kg of water (Ksp=1.78E-5) when [Pb⁺²] = 0.01 M.

Solution:
Step 1) Insert equation of dissociation: PbCl2 > Pb+2 + 2Cl-
Step 2) Insert 1.78E-5 in Ksp-field (Mass solvent = 1 kg)
Step 3) Insert 0.01 in [Pb+2] common ion field
Step 4) Calculate

5.5 The solubility of Ag₂CrO₄ in pure water is 6.54E-5M. Show that the solubility of Ag₂CrO₄ in a 0.05M AgNO₃ sol. is 4.48E-10M.

Solution:
Step 1) Insert equation of dissociation: Ag2CrO4 > 2Ag+ + CrO4-2
Step 2) Insert 6.54E-5 in mol-field (Mass solvent = 1 kg)
Step 3) Calculate solubility product.
Step 4) Insert 0.05 in [Ag+] common ion field
Step 5) Calculate

Back to top

Weak acid/base, buffers

6.1 Calculate [H3O⁺] and pH in a 0.05M HCN^-solution (K_a=5.85E-10).

Solution:
Step 1) Insert equation of dissociation: HCN + H2O > CN- + H3O+
Step 2) Insert (K_a) 5.85E-10
Step 3) Insert [HCN] = 0.05 M
Step 4) Calculate pH and [H3O+]


6.2 Calculate [H3O⁺] and pH pH in a 0.05M acetic acid solution (K_a=1.75E-5). Is acetic acid a stronger acid than HCN ?


Solution:
Step 1) Insert equation of dissociation: CH3COOH + H2O > CH3COO- + H3O+
Step 2) Insert (K_a) = 1.75E-5
Step 3) Insert [CH3COOH] = 0.05 M
Step 4) Calculate pH and [H3O+]
Step 5) Compare acid strength (acetic acid and HCN).


6.3 Calculate K_a for a 0.1 M solution of C₆H₅COOH (benzo acid). ([H3O⁺] = 0.00248 M)

Solution:
Step 1) Insert equation of dissociation: C6H5COOH + H2O > C6H5COO- + H3O+
Step 2) Insert 0.1 in [C6H5COOH] field and 0.00248 in [H3O⁺] field
Step 3) Calculate 6.4 Calculate K_a for a 0.5 M solution of HCOOH (pH = 2.02)

6.4 Calculate K_a in 0.5M HCOOH   (Equlibrium: pH = 2.02).
Solution:
Step 1) Insert equation of dissociation: HCOOH + H2O > HCOO- + H3O+
Step 2) Insert O.5 in [HCOOH] field and 2.02 in pH field
Step 3) Calculate

6.5 Determine [HF] in a pH = 2 solution (K_a=6.94E-4).


Solution:
Step 1) Insert equation of dissociation: HF + H2O > F- + H3O+
Step 2) Insert (Ka)= 6.94E-4
Step 3) Insert pH = 2
Step 4) Calculate
Step 5) Solution: [HF(Before dissoc.)] = [H3O+] + [HF(After dissoc.)].

6.6 Calculate [OH^-] and pH if [NH₃] = 0.05 M (K_b=1.8E-5).


Solution:
Step 1) Insert equation of dissociation: NH3 + H2O > NH4+ + OH-
Step 2) Insert (K_b) = 1.8E-5
Step 3) Insert [NH3] = 0.05 M
Step 4) Calculate pH and [OH-]

6.7 Calculate pH in a 0.1M NH₄Cl solution (KNH4⁺=5.6E-10).

Solution:
NH4Cl > NH4+ + Cl-
Step 1) Insert equation of dissociation: NH4+ + H2O > NH3 + H3O+
Step 2) Insert KNH4+ = 5.6E-10
Step 3) Insert [NH4+] = 0.1 M
Step 4) Calculate pH

6.8 Calculate [OH^-] and pH in a 1 dm3 [NH₃] = 0.2 M (K_b=1.8E-5). Decide pH if 2.675 grams of NH₄Cl is added?

Solution:
Step 1) Insert equation of dissociation: NH3 + H2O > NH4+ + OH-
Step 2) Insert (K_b) = 1.8E-5
Step 3) Insert [NH3] = 0.2 M
Step 4) Calculate pH and [OH-]
Step 5) Calculate (convert) 2.675 g NH4Cl to mole using the molecular calculator: 2.675gNH4Cl = 0.05 mol, in 1 dm3 -> 0.05 M
Step 6)Insert Kb, [NH3] and [NH4+] (common ion field) from NH4Cl
Step 7) Calculate pH

Back to top

Electrochemistry

7.1 What is the overall equation:
Cr₂O₇^-2  +  14H⁺  + 6e^-   -->   2Cr⁺³  +  7H₂O
H₂S   -->   S  +  2H+  +  2e^-


Solution:
Insert following equations in the half cells:
Cr2O7-2  +  14H+ +  6e-  >  2Cr+3  +  7H2O
H2S  >  S  +  2H+  + 2e-
and calculate.

7.2 What is the overall equation:
Cr  -->  Cr⁺³  +  3e^-
MnO4^- +  8H⁺  + e-  -->  Mn⁺² +  4H₂O


Solution:
Insert following half cells reactions:
Cr  >  Cr+3  +  3e-
MnO4- +  8H+  +  e-  >  Mn+2 + 4H2O
and calculate.

7.3 Balance: OH^- + ClO^- + S₂O₃^-2 --> Cl^- + SO₄^-2 + H₂O


Solution:
Insert following equation i one of the half cell fields:
OH- + ClO- + S2O3-2 > Cl- + SO4-2 + H2O
and calculate.

7.4 Dropping a piece of sodium in water. What is the overall reaction. Is this a spontaneous reaction, explain?
2H₂O + 2e^- --> H + 2OH^-       E⁰ = -0.83 V
Na + e --> Na                         E⁰ = -2.71 V
Solution:
Insert following half cells reactions and potentials:
2H2O + 2e-   >  H + 2OH-
Na+  +  e^-     > Na
Remember to turn second half cell reaction.
Calculate.
Spontaneous reaction? -> Investigate sign of overall reaction potential.


7.5 What is the value of the solubility product constant (Ksp) for AgCl?
AgCl  +  e^-   -->  Ag  +  Cl^-       E⁰ = 0.22 V
Ag    -->   Ag⁺ + e^-                 E⁰ = -0.80 V
---------------------------------
AgCl --> Ag⁺² + Cl^-

Solution:
Step 1) Insert first and second half-reaction
AgCl + e- > Ag + Cl-     E⁰ = 0.22 V
Ag > Ag+ + e-                E⁰ = -0.80 V
and calculate equation and overall cell voltage.
Step 2) Insert n=1 (-n=-1) and E⁰  cell=-0.58 in equation: G = -n F E⁰ and calculate.
Step 3) Transfer result of G to equation: -G = R T ln K (remember to change sign of G) and calculate K in which represent the solubility product constant.

Back to top

Spectroscopy

8.1 Use following spectral data to determine the structure of C₇H₈O

MS: 39 53 79 90 107
IR: 700 790
H[1]-NMR: 2.25 7.0
Solution:
Insert spectral data in proper text fields and calculate.
Step 1. Interpretation of common results
MS, IR and H[1]-NMR strongly indicates an aromatic.
MS --> disubst. ; IR (more specific) --> 1,3 disubst.
Step 2. Further investigation
C₇H₈O - C₆H₄ (disubst) = CH₄O (possibly OH and methyl (CH₃))
Verifyed by H[1]-NMR (Ph-CH₃ Methyl H-shift) and MS (CH₂-Ph-OH).
The result:This is a di-subst aromatic compound with a methyl and OH in 1,3 position.

8.2 Determine the structure of C₉H₁₀O₂

MS: 43 65 91
IR: 690 740 1750
H[1]-NMR: 1.95 5.0 7.28


Solution:
Insert spectral data in proper text fields and calculate.
Step 1. Interpretation of common results
MS, IR and H[1]-NMR strongly indicates an aromatic.
MS --> Ph-CH₂ ; IR --> monosubst. benz ; H[1]-NMR --> Aromatic H
What about these oxygens ?:
MS --> CH₃-C=O ; IR --> Esters uconj. C=O & C-O ; H[1]-NMR --> C(=O)-O-CH (esters)
It must be an ester.
Step 2. Further investgation
C₉H₁₀O₂ - phenyl(C₆H₅) = C₃H₅O₂
By MS --> CH₃C=O (MS) and H[1]-NMR --> C(=O)-O-CH , we can conclude that:
C₃H₅O₂ = -CH₂-C(=0)-O-CH₃
The result:This is a mono-subst. aromatic compound: Ph-CH₂-C(=0)-O-CH₃


8.3 Use following spectral data to determine the structure of C₅H₈O₃
MS: 29 45
IR: 1720 3000
H[1]-NMR : 11
C[13]-NMR(multiplicity) : 27.96(t) 29.71(q) 37.83(t) 178.26(s) 207.02(s)


Solution:
Insert spectral data in proper text fields and calculate.
Step 1. Interpretation of common results
MS, IR, H[1]-NMR and C[13]-NMR strongly indicate the attachment of a carboxyl group.
A C=O group (ketone) identifyed by C[13]-NMR (207.07), MS and IR.
Multiplicity: Carbon in both groups = singlets
Step 2. Further investgation
Multiplicity: 27.96(t) = CH₂ 37.83(t) = CH₂ : 29.71(q) = CH₃
The result: CH₃-C(=O)-CH₂-CH₂-C(=O)OH

Back to top

Curve Fit & Function Plot

9.1 The following data describes the relationship between Celsius and Kelvin
Celsius , Kelvin
200 , -73.15
300 , 26.85
400 , 126.85
450 , 176.85
500 , 226.85
What's the equation for this relationship ?
Solution:
Insert the data set in the Data: (X Y) text field and calculate. The equation visible in the Function field represent the relationship.


9.2 How many Fahrenheit is 100 ^oCelsius ?
Celsius , Fahrenheit
-73.15 , -99.67
26.85 , 80.33
126.85 , 260.33
176.85 , 350.33
226.85 , 440.33


Solution:
Insert the data set in the Data: (X Y) text field and calculate. The equation visible in the Function field represent the relationship. Use this equation and calculate Y (X = 100).


9.3 Determine the vapor pressure of butadiene at 0^oC knowing following temperature/pressure relationship.
0^oC , P (mm Hg)
-61.3 , 40
-55.1 , 60
-46.8 , 100
-33.9 , 200
-19.3 , 400
-4.4 , 760
15.3 , 1520
48 , 3700
76.0 , 7600


Solution:
Insert the data set in the Data: (X Y) text field and select Best Fit . The equation visible in the Function field represent the relationship. Use this equation and calculate Y (X = 0).



9.4 Use following experimental data (time and radiation - counts per second) to determine the half life T1/2 for the Pa[234] isotope.
Time(sec) , radiation (cps)
60 , 66.7
150 , 26.9
240 , 11.0
330 , 4.9
420 , 1.7


Solution:
Insert the data set in the Data: (X Y) text field and select Best Fit. The relationship is represented by the equation Y = A * exp(B*X) visible in the Function field. Use the coeff. B in the generated equation and calculate T1/2 knowing that T1/2 = (ln 2)/B .
9.5 The following data where obtained on the rate of hydrolysis of 16% sucrose in 0.1 mol/L HCl aqueous solutions at 34 oC.
t/min , Sucrose remaining %
9.83 , 96.5
59.60 , 80.2
93.18 , 71.0
142.9 , 59.1
294.8 , 32.8
589.4 , 11.1
What is the order of reaction with respect to sucrose ?

Solution:
Insert the data set in the Data: (X Y) text field and select Best Fit. By the curve, an exact exponential decay, we can conclude that the reaction is of first order.

Back to top

Molecular Structure (Molecular 3-D Viewer)

10.1 Examine the stucture of the CO₂ molecule. Is this a polar molecule?
Solution:
Load and examine the structure of CO₂ - co2.pdb.
Keywords: Symmetry


10.2 Examine the stucture of the benzene molecule. Is this a polar molecule? Has benzene a low or high boiling point?


Solution:
Load and examine the structure of benzene - benzene.pdb.
Keywords: Symmetry


10.3 Both benzene and octane contains carbons and hydrogens. What will you expect of the boiling point of a molecule as octane (chain) relatively to a cyclic molecule as benzene.

Solution:
Load and examine the structures of octane and benzene - octane.pdb and benzene.pdb


10.4 What's the reason for the orientation of the H₂O molecules in ice.


Solution:
Examine the molecular orientation in ice - ice.pdb.
Keywords: Polarity, electronegativity, lone pair orbitals, temperature and molecular movements.


10.5 What's d the distance between the atoms in diamond and graphite. What's the distance between the layers in graphite

Solution:
Load and examine the structure of diamond and graphite - diamond.pdb and graphite.pdb
By CHEMIX Dicionary - constants find distances.

10.6 What's the structure of sulfur at boiling point?

Solution:
First, load and examine the structure of sulfur - sulfur8.pdb
At boiling point the sulfur atoms will form S8 rings (single bonds). By increasing this temperature further the sulfur atoms will be broken down to S₆, S₄ and S₂ fragments.

10.7 By word explain the structure of DNA. What kind of atoms are represented and what kind of intermolecular bonds exist in DNA.

Solution:
First, load and examine the structure of a DNA fragment - dna.pdb
It seems like a double helix, and the atoms represented are listed in Statistics text field.

Back to top

Gas Equations

Ideal Gas Law
11.1 How many moles of gas are found in a 1000 dm³ container if the conditions inside the container are 298.15K and 2 atm?
Solution: Insert following values in proper fields: P = 2 atm, T = 298.15 K, V = 1000 dm³ and press Enter.

11.2 What volume will 120 grams of chlorine gas occupy at STP?

Solution:
First, find moles(n) by the use the molecular calculator and argument the Cl₂ by 120g --> 120gCl₂
Second, insert the STP values in proper fields (P,n,T) and press Enter.


11.3A steel tank contains 15.0 g of Cl₂ gas under a pressure of 5.0 atm at 22.0 ^oC. What is the volume of the tank?

Solution:
First, use the units convertor to convert 22 ^oC to Kelvin.
Second, find (n) by inserting 15gCl₂ in the molecular calculator and calculate.
Third, calculate V by inserting the values of P ,n and T in the Ideal Gas Law Calculator.

11.4 A balloon (100 g) was at sea level (1atm and 290K) filled with 1000 dm³ of hydrogen gas. Knowing that air contain approx. 21% oxygen, 78% nitrogen and 1% argon, how many grams was this balloon able to lift?.

Solution:
1) Calculate the mass of 1000 dm³ air:
78% of 1000 dm³ =780 dm³ of N₂ gas.
Insert following values in the Ideal Gas Calculator:
V=790 dm³, P=1atm, T=290K and press Enter.
multiply the result (n) by 28g/mol and find the N₂ mass in the balloon.
Use the same procedure as over, but this time for calculating the masses of O₂ and argon.
2) The mass of 1000 dm³ air is obtained by adding mN₂, mO₂ and argon.
3)Calculate the mass of 1000 dm³ H₂ (V=1000 ,P=1atm T=290K) Multiply the result (n) in this calculation by 2g/mol.
4) Result: mN₂ + mO₂ + mAr - 100g - mH₂ = lift
Combined Gas Law
11.5 A gas was confined in a cylinder fitted with a movable piston. At 290K, the gas occupied a volume of 8.0 dm³ under a pressure of 1.85 atm. The gas was simultaneously heated and compressed, so that its volume was 6.45 dm³ and its temperature was 350K. What pressure was exerted by the hot compressed gas?

Solution:
Insert following values in proper fields: T₁ = 290 K, V₁ = 8.00 dm3, P₁ = 1.85 atm, T₂ = 350 K, V₂ = 6.45 dm³ and press Enter.
Graham's Law of Diffusion
11.6 What is the ratio of the velocity of helium atoms to the velocity of radon atoms (v₁ to v₂) when both gases are at the same temperature? Mass: Radon=222 u, Helium=4.00 u



Solution:
Insert 222 in the m₂ field, 4 in the m₁ field and 1 in the v₂ field. The v₁/v₂ ratio will be calculated by pressing Enter.

Back to top

Conversions

12.1 Convert 298 ^oCelsius and 212 Fahrenheit into Kelvin:
Solution:
Insert given temperature values into proper fields and press Enter.

12.2 Convert the following pressures into units of N/m²:
2 atm, 600 mmHg, 55.21 bar, 30 Torr, 4.8 Pa

Solution:
Insert given values into proper fields and press Enter.
12.3 Convert the following energy-heat units into J (Joule):
23.88 cal, 1 Btu, 0.1 kWh, 1 thermie, 10000000 erg

Solution:
Insert given values into proper fields and press Enter.

Back to top

Ternary Plot

13.1 Four mixtures consisting of chloroform and water was made in different proportions (CH3Cl/H2O):

Describe the changes that occur when fixed amounts of acetic acid are added 5 times to each of the mixtures.
Insert the results in a phase diagram.
Separate 1-phase area and 2-phase area by splines.

Back to top

Back to top

Chemistry Software - Home