First exam Exercises. First Exam/ Exercises 1-Water is an example of: a-a heterogeneous mixture b- a homogeneous mixture c- an element d- a compound 2-The.

First exam Exercises

First Exam/ Exercises 1-Water is an example of: a-a heterogeneous mixture b- a homogeneous mixture c- an element d- a compound 2-The density of a piece of gold with a mass of 301 g and a volume of 15.6 cm 3 is: a-19.3 kg/m 3 b-19.3 g/m 3 c-19.3 g/cm 3 d-19.3 kg/cm 3 d = m/V m = 301 g, V = 15.6 cm 3 d = 301 / 15.6 = 19.3 g/cm 3

First Exam/ Exercises 1-Water is an example of: a-a heterogeneous mixture b- a homogeneous mixture c- an element d- a compound 2-The density of a piece of gold with a mass of 301 g and a volume of 15.6 cm 3 is: a-19.3 kg/m 3 b-19.3 g/m 3 c-19.3 g/cm 3 d-19.3 kg/cm 3 3-SI Base Unit for length is: a- meter b-kilometer c- mile d- foot

First Exam/ Exercises 4-Express 500 centimeters in megameters. a-5.0X10 6 Mm b-5.0X10 -6 Mm c-5.0X10 -8 Mm d-5.0X10 -9 Mm First comvert from cm to m : 500 X 10 -2 m Then from m to Mm 500 x 10 -2 x 10 -6 = 500 x 10 -8 = 5 x10 -6

First Exam/ Exercises 4-Express 500 centimeters in megameters. a-5.0X10 6 Mm b-5.0X10 -6 Mm c-5.0X10 -8 Mm d-5.0X10 -9 Mm 5-Atoms with the same number of protons and with different number of neutrons are called a- ions.b- neutral atoms c- isotopes.d- different atoms. 6-Which of these pairs of elements would be most likely to form an ionic compound? a- P and Brb- Cu and K c- C and O d- O and Zn

First Exam/ Exercises 6-Which of these pairs of elements would be most likely to form an ionic compound? a- P and Brb- Cu and K c- C and O d- O and Zn

First Exam/ Exercises 4-Express 500 centimeters in megameters. a-5.0X10 6 Mm b-5.0X10 -6 Mm c-5.0X10 -8 Mm d-5.0X10 -9 Mm 5-Atoms with the same number of protons and with different number of neutrons are called a- ions.b- neutral atoms c- isotopes. d- different atoms. 6-Which of these pairs of elements would be most likely to form an ionic compound? a- P and Brb- Cu and Kc- C and O d- O and Zn

First Exam/ Exercises 7-How many grams of SF 4 (g) can theoretically be prepared from 6.00 g of SCl 2 (g) and 3.50 g of NaF(s)? The equation of reaction is: 3 SCl 2 (g) + 4 NaF (s) → SF 4 (g) + S2Cl 2 (l) + 4 NaCl (s) a- 21.0 g SF 4 b- 210 g SF 4 c- 2.10 g SF 4 d- 0.210 g SF 4 First we have to determine the limiting reagent: First start with SCl 2 1-Convert to mole : n = 6 / 103 = 0.058 mol 2- from equation 3mole SCl 2 ========= 1 mole SF 4 0.058 mole SCl 2 =====? Mole SF 4 1 x 0.058 = 3 x ? Mole of SF 4 = 0.058 / 3 = 0.019 mole Mass = n x molar mass = 0.019 x 108 =2.052 g ≈ 2.1g second start with NaF 1-Convert to mole : n = 3.5 / 42 = 0.083mol 2- from equation 4mole NaF ========= 1 mole SF 4 0.083 mole SCl 2 =====? Mole SF 4 1 x 0.083 = 4 x ? Mole of SF 4 = 0.083 / 4 = 0.02 mole

First Exam/ Exercises 7-How many grams of SF 4 (g) can theoretically be prepared from 6.00 g of SCl 2 (g) and 3.50 g of NaF(s)? The equation of reaction is: 3 SCl 2 (g) + 4 NaF (s) → SF 4 (g) + S2Cl 2 (l) + 4 NaCl (s) a- 21.0 g SF 4 b- 210 g SF 4 c- 2.10 g SF 4 d- 0.210 g SF 4 8-Calculate the percent composition by mass of C in picric acid (C 6 H 3 N 3 O 7 ). a- 1.3 %b- 18.3 %c- 31.4 % d- 48.9 % Molar mass of C 6 H 3 N 3 O 7 =229g/mol % C= n x molar mass of element molar mass of compound x 100% 6x 12 229 x 100% = 31.4 %

First Exam/ Exercises 7-How many grams of SF 4 (g) can theoretically be prepared from 6.00 g of SCl 2 (g) and 3.50 g of NaF(s)? The equation of reaction is: 3 SCl 2 (g) + 4 NaF (s) → SF 4 (g) + S2Cl 2 (l) + 4 NaCl (s) a- 21.0 g SF 4 b- 210 g SF 4 c- 2.10 g SF 4 d- 0.210 g SF 4 8-Calculate the percent composition by mass of C in picric acid (C 6 H 3 N 3 O 7 ). a- 1.3 %b- 18.3 %c- 31.4 % d- 48.9 % 9-The empirical formula of an organic compound with 85.7% C and 14.3% H is: a- CH b- CH 2 c- C 2 H d- CH 4

First Exam/ Exercises 9-The empirical formula of an organic compound with 85.7% C and 14.3% H is: a- CH b- CH 2 c- C 2 H d- CH 4 1- we change from % to g 85.7 g of C, 14.3 g of H 2- change from g to mole using Divided by the smallest number of mole which is 7.14 85.7 12 = 7.14 mol of C n c = 14.3 1 = 14.3 mol of H n H = 7.14 = 1 C: 7.14 14.3 = 2 H: Thus the empirical formula is CH 2

First Exam/ Exercises 7-How many grams of SF 4 (g) can theoretically be prepared from 6.00 g of SCl 2 (g) and 3.50 g of NaF(s)? The equation of reaction is: 3 SCl 2 (g) + 4 NaF (s) → SF 4 (g) + S2Cl 2 (l) + 4 NaCl (s) a- 21.0 g SF 4 b- 210 g SF 4 c- 2.10 g SF 4 d- 0.210 g SF 4 8-Calculate the percent composition by mass of C in picric acid (C 6 H 3 N 3 O 7 ). a- 1.3 %b- 18.3 %c- 31.4 % d- 48.9 % 9-The empirical formula of an organic compound with 85.7% C and 14.3% H is: a- CH b- CH 2 c- C 2 H d- CH 4

First Exam/ Exercises 10-An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C 6 H 8 O 4 b- C 12 H 16 O 8 c- C 9 H 12 O 6 d- C 15 H 20 O 10 FIRST we calculate the molar mass of emperical formula C 3 H 4 O 2 = 72 g/mol Ratio = 360 / 72 = 5 molecular formula = ratio x empirical formula = 5 x C 3 H 4 O 2 = C 15 H 20 O 10

First Exam/ Exercises 10-An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C 6 H 8 O 4 b- C 12 H 16 O 8 c- C 9 H 12 O 6 d- C 15 H 20 O 10 11-How many grams are there in 3 moles of SF 4 ? a-108 g b- 2.0 x 10 26 g c- 36 g d- 324 g n=mass/molar mass Mass = n x molar mass = 3 x 108 = 324 g From periodic table

First Exam/ Exercises 10-An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C 6 H 8 O 4 b- C 12 H 16 O 8 c- C 9 H 12 O 6 d- C 15 H 20 O 10 11-How many grams are there in 3 moles of SF 4 ? a-108 g b- 2.0 x 10 26 g c- 36 g d- 324 g 12-If you need 1.1 x 10 24 molecules of oxalic acid, H 2 C 2 O 4, how many grams of the acid should you weigh out in the laboratory? 1.826 g b- 0.02 g c- 164.4 g d- 82.2 g

First Exam/ Exercises 12-If you need 1.1 x 10 24 molecules of oxalic acid, H 2 C 2 O 4, how many grams of the acid should you weigh out in the laboratory? a- 1.826 g b- 0.02 g c- 164.4 g d- 82.2 g Number of particale = avogadro number x number of mole 1.1 x10 24 = 6.022 x 10 23 x n n= 1.1 x 10 24 / 6.022 x 10 23 = 1.826 mole n= mass / molar mass Mass = n x molar mass = 1.826 x 90 = 164.4 g

First Exam/ Exercises 10-An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C 6 H 8 O 4 b- C 12 H 16 O 8 c- C 9 H 12 O 6 d- C 15 H 20 O 10 11-How many grams are there in 3 moles of SF 4 ? a-108 g b- 2.0 x 10 26 g c- 36 g d- 324 g 12-If you need 1.1 x 10 24 molecules of oxalic acid, H 2 C 2 O 4, how many grams of the acid should you weigh out in the laboratory? 1.826 g b- 0.02 g c- 164.4 g d- 82.2 g 13-How many grams of Na 2 CO 3 are required for complete reaction with 0.144 g HNO 3 ? Na 2 CO 3 (aq) + 2 HNO 3 (aq) 2 NaNO 3 (aq) + CO 2 (g) + H 2 O(l) a-2.286 x 10 -3 g b- 0.121 g c- 1.143 x 10 -3 g d- 0.072 g

First Exam/ Exercises 13-How many grams of Na 2 CO 3 are required for complete reaction with 0.144 g HNO 3 ? Na 2 CO 3 (aq) + 2 HNO 3 (aq) 2 NaNO 3 (aq) + CO 2 (g) + H 2 O(l) a-2.286 x 10 -3 g b- 0.121 g c- 1.143 x 10 -3 g d- 0.072 g 1-First make sure the equation is balanced 2- g to mole n = mass / molar mass = 0.144 / 63 = 0.0023 mol From equation 2 mole HNO 3 ======= 1 mole Na 2 CO 3 0.002 mole ======== ? mole Na 2 CO 3 1 X 0.002 = 2 X ? mole Na 2 CO 3 = 0.0023 /2 = 0.00115 mole Gram of Na 2 CO 3 = 0.00115 X 106 = 0.121 g X

First Exam/ Exercises 10-An empirical formula of an organic compound is C 3 H 4 O 2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C 6 H 8 O 4 b- C 12 H 16 O 8 c- C 9 H 12 O 6 d- C 15 H 20 O 10 11-How many grams are there in 3 moles of SF 4 ? a-108 g b- 2.0 x 10 26 g c- 36 g d- 324 g 12-If you need 1.1 x 10 24 molecules of oxalic acid, H 2 C 2 O 4, how many grams of the acid should you weigh out in the laboratory? 1.826 g b- 0.02 g c- 164.4 g d- 82.2 g 13-How many grams of Na 2 CO 3 are required for complete reaction with 0.144 g HNO 3 ? Na 2 CO 3 (aq) + 2 HNO 3 (aq) 2 NaNO 3 (aq) + CO 2 (g) + H 2 O(l) a-2.286 x 10 -3 g b- 0.121 g c- 1.143 x 10 -3 g d- 0.072 g

First Exam/ Exercises 14-A glass of water contains 75.96 g of water molecules. How many hydrogen atoms are in the water? 4.57 x 10 25 b- 5.08 x 10 24 c- 2.54 x 10 24 d- 9.15 x 10 25 First we calculate the number of mole n = 75.96 / 18 = 4.22 mole Number of molecules = Avogadro's number x number of mole = 6.022 x 10 23 x 4.22 = 2.54 x10 24 molecules From the chemical formula of water H 2 O 1 molecules of water = 2 atom of H 2.539 x 10 24 molecules = ? Atom of H 2 x 2.54 x 10 24 = 5.08 x10 24 atoms

First Exam/ Exercises 14-A glass of water contains 75.96 g of water molecules. How many hydrogen atoms are in the water? 4.57 x 10 25 b- 5.08 x 10 24 c- 2.54 x 10 24 d- 9.15 x 10 25 15- has the correct set of (Z = Atomic number, A = mass number, c = charge ) a- Z = 16, A = 8, c = 0 b- Z = 8, A = 16, c = 1 c- Z = 8, A = 16, c = -2 d- Z = 8, A = 16, c = 0 16-Phosphorus ion (P -3 ) with (Z=15 and A= 31) has the correct set of (e = electrons and n = neutrons) a- e = 18, n = 16 b- e = 15, n = 16 c- e = 15, n = 18 d- e = 16, n = 18 P = 15, e = 15+3 = 18, n = 31-15 = 16

First Exam/ Exercises 14-A glass of water contains 75.96 g of water molecules. How many hydrogen atoms are in the water? 4.57 x 10 25 b- 5.08 x 10 24 c- 2.54 x 10 24 d- 9.15 x 10 25 15- has the correct set of (Z = Atomic number, A = mass number, c = charge ) a- Z = 16, A = 8, c = 0 b- Z = 8, A = 16, c = 1 c- Z = 8, A = 16, c = -2 d- Z = 8, A = 16, c = 0 16-Phosphorus ion (P -3 ) with (Z=15 and A= 31) has the correct set of (e = electrons and n = neutrons) a- e = 18, n = 16 b- e = 15, n = 16 c- e = 15, n = 18 d- e = 16, n = 18

First Exam/ Exercises 17-Vanadium (V) has two stable isotopes, ( 50 V and 51 V). The average atomic mass of Vanadium (V) is 50.9415 amu; 50 V has a mass of 49.9462 amu with an abundance of 0.25%. The mass of ( 51 V) isotope in amu with an abundance of 99.75% is a- 51.4940 b- 50.9440 c- 50. 4490 d- 50.9404 Average atomic mass = (atomic mass x abundenace) V-50 + (atomic mass x abundenace) V-51 50.9415 = (49.9462 X (0.25 /100)) V-50 + ( atomic mass x (99.75 /100)) V51 50.9415 = 0.125 + (ATMOIC MASS OF V-51 X 0.9975) 50.9415 – 0.125 = (ATMOIC MASS OF V-51 X 0.9975) 50.8166=(ATMOIC MASS OF V-51 X 0.9975) ATOMIC MASS OF V-51 = 50.8166 / 0.9975= 50.944

First Exam/ Exercises 17-Vanadium (V) has two stable isotopes, ( 50 V and 51 V). The average atomic mass of Vanadium (V) is 50.9415 amu; 50 V has a mass of 49.9462 amu with an abundance of 0.25%. The mass of ( 51 V) isotope in amu with an abundance of 99.75% is a- 51.4940 b- 50.9440 c- 50. 4490 d- 50.9404 18-The element would be most likely to form cation is a- O b- He c- F d- Pb

First Exam/ Exercises 18-The element would be most likely to form cation is a- O b- He c- F d- Pb

First Exam/ Exercises 17-Vanadium (V) has two stable isotopes, ( 50 V and 51 V). The average atomic mass of Vanadium (V) is 50.9415 amu; 50 V has a mass of 49.9462 amu with an abundance of 0.25%. The mass of ( 51 V) isotope in amu with an abundance of 99.75% is a- 51.4940 b- 50.9440 c- 50. 4490 d- 50.9404 18-The element would be most likely to form cation is a- O b- He c- F d- Pb 19-The element in group 7A and period 6 is a- At b- Po c- I d- Rn 20-The empirical formula of C 6 H 8 O 6 is a- C 6 H 8 O 6 b- C 3 H 4 O 3 c- C 2 H 4 O 2 d- C 2 H 4 O Dived each by 2

First Exam/ Exercises 17-Vanadium (V) has two stable isotopes, ( 50 V and 51 V). The average atomic mass of Vanadium (V) is 50.9415 amu; 50 V has a mass of 49.9462 amu with an abundance of 0.25%. The mass of ( 51 V) isotope in amu with an abundance of 99.75% is a- 51.4940 b- 50.9440 c- 50. 4490 d- 50.9404 18-The element would be most likely to form cation is a- O b- He c- F d- Pb 19-The element in group 7A and period 6 is a- At b- Po c- I d- Rn 20-The empirical formula of C 6 H 8 O 6 is a- C 6 H 8 O 6 b- C 3 H 4 O 3 c- C 2 H 4 O 2 d- C 2 H 4 O

First Exam/ Exercises 21-The correct name for N 2 O 3 is a-Dinitrogen monoxide b- Dinitrogen trioxide c- Nitrogen dioxid d- Dinitrogen pentaoxide

First Exam/ Exercises Summery of naming compound IonicMolecular Cation: metal or NH 4 + Anion: monotomic or polytomic Cation has only one charge Cation has more than one charge Alkali metal Alkaline earth metal Ag +, Al +3, Cd +2, Zn +2 Other metal cations Name metal first If monoatomic anion, add ide to the anion If polyatomic anion use name of anion from previous table Name metal first Specify charge of metal cation with roman numeral (STOCK SYSTEM) If monoatomic anion, add ide to the anion If polyatomic anion use name of anion from previous table Nonmetal + nonmetal Nonmetal + metalloid Pair Form one type of compound Pair Form more than one type of compound Name first element add ide to the name of second element Name first element add ide to the name of second element Add the prefix (prefix mono usually omitted for the first element

First Exam/ Exercises 21-The correct name for N 2 O 3 is a-Dinitrogen monoxide b- Dinitrogen trioxide c- Nitrogen dioxid d- Dinitrogen pentaoxide 22-The systemic name of Cu 3 (PO 4 ) 2 is a- Cuprous phosphate b- Copper (II) phosphate c- Copper (I) phosphate d- Copper (III) phosphate 23-ratio of protons to electrons in (O -2 ) with (Z=8 and A= 16) is a- 1:2 b- 2:1 c- 4:5 d- 1:1 P =8, e = 8+2=10 P:e 8:10 4:5

First Exam/ Exercises 21-The correct name for N 2 O 3 is a-Dinitrogen monoxide b- Dinitrogen trioxide c- Nitrogen dioxid d- Dinitrogen pentaoxide 22-The systemic name of Cu 3 (PO 4 ) 2 is a- Cuprous phosphate b- Copper (II) phosphate c- Copper (I) phosphate d- Copper (III) phosphate 23-ratio of protons to electrons in (O -2 ) with (Z=8 and A= 16) is a- 1:2 b- 2:1 c- 4:5 d- 1:1 24-How many moles of MgCl 2 are present in 60.0 mL of 0.100 M MgCl 2 solution? a-60.0 mol b- 0.572 mol c- 6.00 × 10 -3 mol d- 6.00 mol M= 0.100M, n=?, V= 60 mL= 60/1000=0.06L M=n/V n=M X V = 0.1 X 0.06 = 0.006 mol = 6 x 10 -3 mol

First Exam/ Exercises 21-The correct name for N 2 O 3 is a-Dinitrogen monoxide b- Dinitrogen trioxide c- Nitrogen dioxid d- Dinitrogen pentaoxide 22-The systemic name of Cu 3 (PO 4 ) 2 is a- Cuprous phosphate b- Copper (II) phosphate c- Copper (I) phosphate d- Copper (III) phosphate 23-ratio of protons to electrons in (O -2 ) with (Z=8 and A= 16) is a- 1:2 b- 2:1 c- 4:5 d- 1:1 24-How many moles of MgCl 2 are present in 60.0 mL of 0.100 M MgCl 2 solution? a-60.0 mol b- 0.572 mol c- 6.00 × 10 -3 mol d- 6.00 mol

First Exam/ Exercises 25-Which of the following solutions contains the largest number of moles of solute? a-26.5 mL of 4.9 M sodium chloride b- 45 mL of 0.99 M hydrochloric acid c- 80 mL of 0.5 M nitric acid d- 100 mL of 1 M sodium hydroxide a- n= 4.9 x 26.5/1000 = 0.13 mol b- n= 0.99 x 45/1000=0.044 mol c- n= 0.5 x80/1000 = 0.04 mol d- n= 1 x 100/1000=0.1 mol

First Exam/ Exercises 25-Which of the following solutions contains the largest number of moles of solute? a-26.5 mL of 4.9 M sodium chloride b- 45 mL of 0.99 M hydrochloric acid c- 80 mL of 0.5 M nitric acid d- 100 mL of 1 M sodium hydroxide 26-Calculate the molarity of a solution of 2.50 g of ethanol (C 2 H 5 OH) in 545 mL of solution. a-4.59 × 10 -3 M b- 9.96 × 10 -2 M c- 1.36 M d- 0.218 M M = n/V n = mass /molar mass = 2.5 / 46= 0.0543 mole M = 0.0543 / (545 / 1000)= 0.0996 M = 9.96 X 10 -2 M

First Exam/ Exercises 25-Which of the following solutions contains the largest number of moles of solute? a-26.5 mL of 4.9 M sodium chloride b- 45 mL of 0.99 M hydrochloric acid c- 80 mL of 0.5 M nitric acid d- 100 mL of 1 M sodium hydroxide 26-Calculate the molarity of a solution of 2.50 g of ethanol (C 2 H 5 OH) in 545 mL of solution. a-4.59 × 10 -3 M b- 9.96 × 10 -2 M c- 1.36 M d- 0.218 M 27-Calculate the mass of NaNO 3 in grams required to prepare 250 mL of a 0.173 M solution. a-3.68 g b- 15.0 g c- 43.3 g d- 3.68 × 10 3 g

First Exam/ Exercises 27-Calculate the mass of NaNO 3 in grams required to prepare 250 mL of a 0.173 M solution. a-3.68 g b- 15.0 g c- 43.3 g d- 3.68 × 10 3 g M = n/V n = MXV = 0.173 X (250 /1000)= 0.04325 mol Mass = n x molar mass = 0.04325 x 85 = 3.68 g

First Exam/ Exercises 25-Which of the following solutions contains the largest number of moles of solute? a-26.5 mL of 4.9 M sodium chloride b- 45 mL of 0.99 M hydrochloric acid c- 80 mL of 0.5 M nitric acid d- 100 mL of 1 M sodium hydroxide 26-Calculate the molarity of a solution of 2.50 g of ethanol (C 2 H 5 OH) in 545 mL of solution. a-4.59 × 10 -3 M b- 9.96 × 10 -2 M c- 1.36 M d- 0.218 M 27-Calculate the mass of NaNO 3 in grams required to prepare 250 mL of a 0.173 M solution. a-3.68 g b- 15.0 g c- 43.3 g d- 3.68 × 10 3 g 28-A 10.0 mL sample of 16.5 M HF is diluted to a final volume of 250 mL. What is the molarity of the final solution? a-1.52 M b- 0.660 M c- 0.100 M d- 1.06 M M 1 V 1 = M 2 V 2 16.5 X 10 = M 2 X 250 M 2 = 16.5 X 10 / 250 = 0.66 M

First Exam/ Exercises 25-Which of the following solutions contains the largest number of moles of solute? a-26.5 mL of 4.9 M sodium chloride b- 45 mL of 0.99 M hydrochloric acid c- 80 mL of 0.5 M nitric acid d- 100 mL of 1 M sodium hydroxide 26-Calculate the molarity of a solution of 2.50 g of ethanol (C 2 H 5 OH) in 545 mL of solution. a-4.59 × 10 -3 M b- 9.96 × 10 -2 M c- 1.36 M d- 0.218 M 27-Calculate the mass of NaNO 3 in grams required to prepare 250 mL of a 0.173 M solution. a-3.68 g b- 15.0 g c- 43.3 g d- 3.68 × 10 3 g 28-A 10.0 mL sample of 16.5 M HF is diluted to a final volume of 250 mL. What is the molarity of the final solution? a-1.52 M b- 0.660 M c- 0.100 M d- 1.06 M

First Exam/ Exercises 29-A 10.0 mL of 0.665 M KMnO 4 solution is mixed with 16.7 mL of 0.892 M KMnO 4 solution. Calculate the concentration of the final solution. a-0.778 M b- 0.807 M c- 2.37 M d- 0.411 M For the first solution n = MXV = 0.665 X (10/1000) = 0.00665 mol For the second solution n = MX V = 0.892 X (16.7 /1000) = 0.014896 n for the total n = 0.00665 + 0.014896 = 0.02154 mol V for total V= 10 + 16.7 = 26.7 ml = 0.0267L M = n/V = 0.02154 / 0.0267 = 0.807M

First Exam/ Exercises 29-A 10.0 mL of 0.665 M KMnO 4 solution is mixed with 16.7 mL of 0.892 M KMnO 4 solution. Calculate the concentration of the final solution. a-0.778 M b- 0.807 M c- 2.37 M d- 0.411 M 30-The following reaction describes combustion of an alkane, the correct balance will be with a C 5 H 12 + b O 2 → c CO 2 + d H 2 O a- a = 5, b = 2, c = 2, d = 2b- a = 12, b = 2, c = 2, d = 2 c- a = 1, b = 8, c = 5, d = 6d- a = 1, b = 4, c = 5, d = 6

First Exam/ Exercises 30-The following reaction describes combustion of an alkane, the correct balance will be with a C 5 H 12 + b O 2 → c CO 2 + d H 2 O a- a = 5, b = 2, c = 2, d = 2b- a = 12, b = 2, c = 2, d = 2 c- a = 1, b = 8, c = 5, d = 6d- a = 1, b = 4, c = 5, d = 6 C 5 H 12 + O 2 → CO 2 + H 2 O C 5 C 1 X5 C 5 H 12 + O 2 → 5 CO 2 + H 2 O H 12 H2 X6 C 5 H 12 + O 2 → 5 CO 2 + 6H 2 O O 2 O 10 + 6 =16 X 8 C 5 H 12 + 8 O 2 → 5 CO 2 + 6H 2 O

First Exam/ Exercises 29-A 10.0 mL of 0.665 M KMnO 4 solution is mixed with 16.7 mL of 0.892 M KMnO 4 solution. Calculate the concentration of the final solution. a-0.778 M b- 0.807 M c- 2.37 M d- 0.411 M 30-The following reaction describes combustion of an alkane, the correct balance will be with a C 5 H 12 + b O 2 → c CO 2 + d H 2 O a- a = 5, b = 2, c = 2, d = 2b- a = 12, b = 2, c = 2, d = 2 c- a = 1, b = 8, c = 5, d = 6d- a = 1, b = 4, c = 5, d = 6

First exam Exercises. First Exam/ Exercises 1-Water is an example of: a-a heterogeneous mixture b- a homogeneous mixture c- an element d- a compound 2-The.

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First exam Exercises. First Exam/ Exercises 1-Water is an example of: a-a heterogeneous mixture b- a homogeneous mixture c- an element d- a compound 2-The.

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