First exam Exercises.

First exam Exercises

1-Water is an example of:
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 cm3 is: a-19.3 kg/m b-19.3 g/m c-19.3 g/cm d-19.3 kg/cm3 d = m/V m = 301 g, V = 15.6 cm3 d = 301 / 15.6 = 19.3 g/cm3

1-Water is an example of:
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 cm3 is: a-19.3 kg/m b-19.3 g/m c-19.3 g/cm d-19.3 kg/cm3 3-SI Base Unit for length is: a- meter b-kilometer c- mile d- foot

4-Express 500 centimeters in megameters.
First Exam/ Exercises 4-Express 500 centimeters in megameters. a-5.0X106 Mm b-5.0X10-6 Mm c-5.0X10-8 Mm d-5.0X10-9 Mm First comvert from cm to m : 500 X 10-2m 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-5.0X106 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 Br b- Cu and K c- C and O d- O and Zn

a- P and Br b- 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 Br b- Cu and K c- C and O d- O and Zn

First Exam/ Exercises 4-Express 500 centimeters in megameters. a-5.0X106 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 Br b- Cu and K c- C and O d- O and Zn

3 SCl2 (g) + 4 NaF (s) → SF4 (g) + S2Cl2 (l) + 4 NaCl (s)
First Exam/ Exercises 7-How many grams of SF4 (g) can theoretically be prepared from 6.00 g of SCl2 (g) and 3.50 g of NaF(s)? The equation of reaction is: 3 SCl2 (g) NaF (s) → SF4 (g) + S2Cl2 (l) NaCl (s) a g SF4 b- 210 g SF c g SF4 d g SF4 First we have to determine the limiting reagent: second start with NaF 1-Convert to mole : n = 3.5 / 42 = 0.083mol 2- from equation 4mole NaF ========= 1 mole SF4 0.083 mole SCl2 =====? Mole SF4 1 x = 4 x ? Mole of SF4 = / 4 = 0.02 mole First start with SCl2 1-Convert to mole : n = 6 / 103 = mol 2- from equation 3mole SCl2 ========= 1 mole SF4 0.058 mole SCl2 =====? Mole SF4 1 x = 3 x ? Mole of SF4 = / 3 = mole Mass = n x molar mass = x 108 =2.052 g ≈ 2.1g

First Exam/ Exercises 7-How many grams of SF4 (g) can theoretically be prepared from 6.00 g of SCl2 (g) and 3.50 g of NaF(s)? The equation of reaction is: 3 SCl2 (g) NaF (s) → SF4 (g) + S2Cl2 (l) NaCl (s) a g SF4 b- 210 g SF c g SF4 d g SF4 8-Calculate the percent composition by mass of C in picric acid (C6H3N3O7). a % b % c % d % Molar mass of C6H3N3O7 =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 SF4 (g) can theoretically be prepared from 6.00 g of SCl2 (g) and 3.50 g of NaF(s)? The equation of reaction is: 3 SCl2 (g) NaF (s) → SF4 (g) + S2Cl2 (l) NaCl (s) a g SF4 b- 210 g SF c g SF4 d g SF4 8-Calculate the percent composition by mass of C in picric acid (C6H3N3O7). a % b % c % d % 9-The empirical formula of an organic compound with 85.7% C and 14.3% H is: a- CH b- CH2 c- C2H d- CH4

2- change from g to mole using
First Exam/ Exercises 9-The empirical formula of an organic compound with 85.7% C and 14.3% H is: a- CH b- CH2 c- C2H d- CH4 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 Thus the empirical formula is CH2 85.7 12 = 7.14 mol of C nc = 14.3 1 = 14.3 mol of H nH = 7.14 = 1 7.14 14.3 = 2 H: C:

First Exam/ Exercises 7-How many grams of SF4 (g) can theoretically be prepared from 6.00 g of SCl2 (g) and 3.50 g of NaF(s)? The equation of reaction is: 3 SCl2 (g) NaF (s) → SF4 (g) + S2Cl2 (l) NaCl (s) a g SF4 b- 210 g SF c g SF4 d g SF4 8-Calculate the percent composition by mass of C in picric acid (C6H3N3O7). a % b % c % d % 9-The empirical formula of an organic compound with 85.7% C and 14.3% H is: a- CH b- CH2 c- C2H d- CH4

a- C6H8O4 b- C12H16O8 c- C9H12O6 d- C15H20O10
First Exam/ Exercises 10-An empirical formula of an organic compound is C3H4O2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C6H8O b- C12H16O8 c- C9H12O6 d- C15H20O10 FIRST we calculate the molar mass of emperical formula C3H4O2 = 72 g/mol Ratio = 360 / 72 = 5 molecular formula = ratio x empirical formula = 5 x C3H4O2 = C15H20O10

First Exam/ Exercises 10-An empirical formula of an organic compound is C3H4O2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C6H8O b- C12H16O8 c- C9H12O6 d- C15H20O10 11-How many grams are there in 3 moles of SF4? a-108 g b- 2.0 x 1026 g c- 36 g d- 324 g n=mass/molar mass SF4= 108 g/mol Mass = n x molar mass = 3 x 108 = 324 g

First Exam/ Exercises 10-An empirical formula of an organic compound is C3H4O2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C6H8O b- C12H16O8 c- C9H12O6 d- C15H20O10 11-How many grams are there in 3 moles of SF4? a-108 g b- 2.0 x 1026 g c- 36 g d- 324 g 12-If you need 1.1 x 1024 molecules of oxalic acid, H2C2O4, how many grams of the acid should you weigh out in the laboratory? 1.826 g b g c g d g

Number of particle = Avogadro number x number of mole
First Exam/ Exercises 12-If you need 1.1 x 1024 molecules of oxalic acid, H2C2O4, how many grams of the acid should you weigh out in the laboratory? a g b g c g d g Number of particle = Avogadro number x number of mole 1.1 x1024 = x 1023 x n n= 1.1 x 1024 / x 1023 = mole n= mass / molar mass Mass = n x molar mass = x 90 = g

Na2CO3(aq) + 2 HNO3(aq) 2 NaNO3(aq) + CO2(g) + H2O(l)
First Exam/ Exercises 10-An empirical formula of an organic compound is C3H4O2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C6H8O b- C12H16O8 c- C9H12O6 d- C15H20O10 11-How many grams are there in 3 moles of SF4? a-108 g b- 2.0 x 1026 g c- 36 g d- 324 g 12-If you need 1.1 x 1024 molecules of oxalic acid, H2C2O4, how many grams of the acid should you weigh out in the laboratory? 1.826 g b g c g d g 13-How many grams of Na2CO3 are required for complete reaction with g HNO3? Na2CO3(aq) + 2 HNO3(aq) NaNO3(aq) + CO2(g) + H2O(l) a x 10-3 g b g c x 10-3 g d g

First Exam/ Exercises 13-How many grams of Na2CO3 are required for complete reaction with g HNO3? Na2CO3(aq) + 2 HNO3(aq) NaNO3(aq) + CO2(g) + H2O(l) a x 10-3 g b g c x 10-3 g d g 1-First make sure the equation is balanced 2- g to mole n = mass / molar mass = / 63 = mol From equation 2 mole HNO3 ======= 1 mole Na2CO3 mole ======== ? mole Na2CO3 1 X = 2 X ? mole Na2CO3 = /2 = mole Gram of Na2CO3 = X 106 = g X

First Exam/ Exercises 10-An empirical formula of an organic compound is C3H4O2, if the molecular weight of the compound is (360 g/mol), the molecular formula of the compound will be: a- C6H8O b- C12H16O8 c- C9H12O6 d- C15H20O10 11-How many grams are there in 3 moles of SF4? a-108 g b- 2.0 x 1026 g c- 36 g d- 324 g 12-If you need 1.1 x 1024 molecules of oxalic acid, H2C2O4, how many grams of the acid should you weigh out in the laboratory? 1.826 g b g c g d g 13-How many grams of Na2CO3 are required for complete reaction with g HNO3? Na2CO3(aq) + 2 HNO3(aq) NaNO3(aq) + CO2(g) + H2O(l) a x 10-3 g b g c x 10-3 g d g

First we calculate the number of mole n = 75.96 / 18 = 4.22 mole
First Exam/ Exercises 14-A glass of water contains g of water molecules. How many hydrogen atoms are in the water? 4.57 x b x c x d x 1025 First we calculate the number of mole n = / 18 = 4.22 mole Number of molecules = Avogadro's number x number of mole = x 1023 x 4.22 = 2.54 x1024 molecules From the chemical formula of water H2O 1 molecules of water = 2 atom of H 2.539 x 1024 molecules = ? Atom of H 2 x 2.54 x 1024 = 5.08 x1024 atoms

First Exam/ Exercises 14-A glass of water contains g of water molecules. How many hydrogen atoms are in the water? 4.57 x b x c x d x 1025 has the correct set of (Z = Atomic number, A = mass number, c = charge ) a- Z = 16, A = 8, c = b- Z = 8, A = 16, c = 1 c- Z = 8, A = 16, c = 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 = b e = 15, n = 16 c- e = 15, n = d e = 16, n = 18 P = 15, e = 15+3 = 18 , n = = 16

First Exam/ Exercises 14-A glass of water contains g of water molecules. How many hydrogen atoms are in the water? 4.57 x b x c x d x 1025 has the correct set of (Z = Atomic number, A = mass number, c = charge ) a- Z = 16, A = 8, c = b- Z = 8, A = 16, c = 1 c- Z = 8, A = 16, c = 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 = b e = 15, n = 16 c- e = 15, n = d e = 16, n = 18

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

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

18-The element would be most likely to form cation is
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, (50V and 51V). The average atomic mass of Vanadium (V) is amu; 50V has a mass of amu with an abundance of 0.25%. The mass of (51V) isotope in amu with an abundance of 99.75% is a b c d 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 C6H8O6 is a- C6H8O b- C3H4O c- C2H4O d- C2H4O Dived each by 2

First Exam/ Exercises 17-Vanadium (V) has two stable isotopes, (50V and 51V). The average atomic mass of Vanadium (V) is amu; 50V has a mass of amu with an abundance of 0.25%. The mass of (51V) isotope in amu with an abundance of 99.75% is a b c d 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 C6H8O6 is a- C6H8O b- C3H4O c- C2H4O d- C2H4O

21-The correct name for N2O3 is
First Exam/ Exercises 21-The correct name for N2O3 is a-Dinitrogen monoxide b- Dinitrogen trioxide c- Nitrogen dioxid d- Dinitrogen pentaoxide

First Exam/ Exercises Summery of naming compound Ionic Molecular
Cation: metal or NH4+ Anion: monotomic or polytomic Nonmetal + nonmetal Nonmetal + metalloid Cation has only one charge Cation has more than one charge 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 Alkali metal Alkaline earth metal Ag+, Al+3, Cd+2, Zn+2 Other metal cations Name first element add ide to the name of second element Add the prefix (prefix mono usually omitted for the first element 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 Name metal first If monoatomic anion, add ide to the anion If polyatomic anion use name of anion from previous table

First Exam/ Exercises 21-The correct name for N2O3 is a-Dinitrogen monoxide b- Dinitrogen trioxide c- Nitrogen dioxid d- Dinitrogen pentaoxide 22-The systemic name of Cu3(PO4)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: b- 2: c- 4: d- 1:1 P =8 , e = 8+2=10 P:e 8:10 4:5

First Exam/ Exercises 21-The correct name for N2O3 is a-Dinitrogen monoxide b- Dinitrogen trioxide c- Nitrogen dioxid d- Dinitrogen pentaoxide 22-The systemic name of Cu3(PO4)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: b- 2: c- 4: d- 1:1 24-How many moles of MgCl2 are present in 60.0 mL of M MgCl2 solution? a-60.0 mol b mol c × 10-3 mol d mol M= 0.100M, n=?, V= 60 mL= 60/1000=0.06L M=n/V n=M X V = 0.1 X 0.06 = mol = 6 x 10-3 mol

First Exam/ Exercises 21-The correct name for N2O3 is a-Dinitrogen monoxide b- Dinitrogen trioxide c- Nitrogen dioxid d- Dinitrogen pentaoxide 22-The systemic name of Cu3(PO4)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: b- 2: c- 4: d- 1:1 24-How many moles of MgCl2 are present in 60.0 mL of M MgCl2 solution? a-60.0 mol b mol c × 10-3 mol d mol

c- 80 mL of 0.5 M nitric acid d- 100 mL of 1 M sodium hydroxide
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 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 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 (C2H5OH) in 545 mL of solution. a-4.59 × 10-3 M b × 10-2 M c M d M M = n/V n = mass /molar mass = 2.5 / 46= mole M = / (545 / 1000)= M = 9.96 X 10-2M

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 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 (C2H5OH) in 545 mL of solution. a-4.59 × 10-3 M b × 10-2 M c M d M 27-Calculate the mass of NaNO3 in grams required to prepare 250 mL of a M solution. a-3.68 g b g c g d × 103 g

First Exam/ Exercises 27-Calculate the mass of NaNO3 in grams required to prepare 250 mL of a M solution. a-3.68 g b g c g d × 103 g M = n/V n = MXV = X (250 /1000)= mol Mass = n x molar mass = 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 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 (C2H5OH) in 545 mL of solution. a-4.59 × 10-3 M b × 10-2 M c M d M 27-Calculate the mass of NaNO3 in grams required to prepare 250 mL of a M solution. a-3.68 g b g c g d × 103 g

First Exam/ Exercises 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 M c M d M M1 V1 = M2 V2 16.5 X 10 = M2 X 250 M2= 16.5 X 10 / 250 = 0.66 M

First Exam/ Exercises 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 M c M d M 29-A 10.0 mL of M KMnO4 solution is mixed with 16.7 mL of M KMnO4 solution. Calculate the concentration of the final solution. a M b M c M d M

For the second solution n = MX V = 0.892 X (16.7 /1000) = 0.014896
First Exam/ Exercises 29-A 10.0 mL of M KMnO4 solution is mixed with 16.7 mL of M KMnO4 solution. Calculate the concentration of the final solution. a M b M c M d M For the first solution n = MXV = X (10/1000) = mol For the second solution n = MX V = X (16.7 /1000) = n for the total n = = mol V for total V= = 26.7 ml = L M = n/V = / = 0.807M

a- a = 5, b = 2, c = 2, d = 2 b- a = 12, b = 2, c = 2, d = 2
First Exam/ Exercises 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 M c M d M 29-A 10.0 mL of M KMnO4 solution is mixed with 16.7 mL of M KMnO4 solution. Calculate the concentration of the final solution. a M b M c M d M 30-The following reaction describes combustion of an alkane, the correct balance will be with a C5H12 + b O2 → c CO2 + d H2O a- a = 5, b = 2, c = 2, d = 2 b- a = 12, b = 2, c = 2, d = 2 c- a = 1, b = 8, c = 5, d = 6 d- a = 1, b = 4, c = 5, d = 6

a- a = 5, b = 2, c = 2, d = 2 b- a = 12, b = 2, c = 2, d = 2
First Exam/ Exercises 30-The following reaction describes combustion of an alkane, the correct balance will be with a C5H12 + b O2 → c CO2 + d H2O a- a = 5, b = 2, c = 2, d = 2 b- a = 12, b = 2, c = 2, d = 2 c- a = 1, b = 8, c = 5, d = 6 d- a = 1, b = 4, c = 5, d = 6 C5H12 + O2 → CO2 + H2O C C X5 C5H12 + O2 → 5 CO2 + H2O H H X6 C5H12 + O2 → 5 CO2 + 6H2O O O = X 8 C5H O2 → 5 CO2 + 6H2O

a- a = 5, b = 2, c = 2, d = 2 b- a = 12, b = 2, c = 2, d = 2
First Exam/ Exercises 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 M c M d M 29-A 10.0 mL of M KMnO4 solution is mixed with 16.7 mL of M KMnO4 solution. Calculate the concentration of the final solution. a M b M c M d M 30-The following reaction describes combustion of an alkane, the correct balance will be with a C5H12 + b O2 → c CO2 + d H2O a- a = 5, b = 2, c = 2, d = 2 b- a = 12, b = 2, c = 2, d = 2 c- a = 1, b = 8, c = 5, d = 6 d- a = 1, b = 4, c = 5, d = 6

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