نموذج c كيمياء من الشابردي

[شابردي]

يا حلوين هذا النموذج الثالث و الله اني راح انام عليه

Exam 1(C) Choose and mark the correct answer : C1-The letter c in the chemical symbol ( ( ) ) ) represents: a-Atomic number b-mass number c-charge d- frequency Answer: C C2- Mercury is an example of a a-Compound b- element c- homogeneous mixture d- heterogeneous mixture Answer: B C3- The unit of speed (c) is a- m s-2 b-kg m2 s-2 c- m s-1 d-kg m s-2 Answer: c C4-1.27 cm is equal to a-1.27 × 10-10m b-1.27×10-3 mm c-1.27×10+3pm d-1.27×10-2m Answer: D C5-The number (0.0099899) written to one significant Figures is a- 0.00999 b-0.009990 c- 1.0 ×10-2 d-1×10-2 Answer: c C6-The correct Significant Figures result of the following operation () is a-one b- two c- three d- four Answer: B C7- A particular medicine dropper delivers 22 drops of water to make 1.0 mL. The volume of three drops in cm3 is a-0.045 cm3 b-0.091 cm3 c-0.136 cm3 d- 0.182 cm3 Answer: C C8-For which of the following calculations is 3.2×10-1 the correct answer? a-145.75+(2.3×10-1) b-79.500÷(2.5×102) c-(7.0×10-3)-(8.0×10-4) d-(1.0×104) ×(9.9×106) Answer: B C9-Cesium atoms are the largest naturally occurring atom . The radius of cesium atom is 2.62 A. How many cesium atoms would have to be laid side by side to give a row of cesium atoms 2.0 cm long ? Assume the atoms are spherical a-4.85×107 b-1.91×107 c-3.82 ×107 d-5.73×107 Answer: D C10-The international prototype of the kilogram is a cylinder made from an alloy that is 90.000% platinum and 10.000 % iridium. How many atoms of PT are in the cylinder ? a-4.62 b-0.52 c-3.133×1023 d- 2.78×1024 Answer: D C11-The number of neutrons (n) and electron (e-) in the element zinc () is a-30 p, 28 e b-35 n , 30 p c-31 n , 30 p d-35 n ,28 p Answer: D C12- Silver occurs in nature as mixture of two isotopes : and .The number of neutrons in both isotopes ( and ) is a- The same b- bigger in isotopes c- smaller in isotopes d-none of the above Answer: A C13- How many moles of sulfur are there in 9.03 ×1023 sulfur – atoms ? a-0.05 mol b-1.0 mol c 1.5 mol d-2.0 mol Answer: C C14-Sodium hydrogen carbonate , commonly called “bicarbonate of soda “ is used in many commercial products to relieve an upset stomach .It has the simplest formula NaHCO3 . The percentage composition of carbon is a—27.38% b-1.20% c-14.29% d-57.14% Answer: C C15-For the reaction: 2 PbS (s) +3 O2 (g) → 2 PbO (s) + 2 SO2 (g) If 0.0252 mol of SO2 reacts with oxygen , how many moles of PbS are produced? a-0.0252 b-7.95 c 1.35 d- 5.30 Answer: A C16-Naphthalene , contains only carbon and hydrogen .Combustion of a 1.000 g sample of naphthalene gives 0.562 g of water H2O .what is the number of moles of hydrogen in 1.000 g naphthalene ? a-6.24% b-93.7% c-0.0642 d-0.078 Answer: C C17-The molecular of weight of the empirical formula of compound (C3H4O2) is (72 g⁄mol) if the molecular weight of the compound is (144 g⁄mol) , the molecular formula of the compound is a-C6H8O4 b- C12H16O8 c-C9H12O6 d-C18H24O12 Answer: A C18-The number of oxygen-atoms in (0.5 mol) of the aminoacid alanine (CH3CHNH2COOH) is a-9.02×1023 b-2.11×1024 c-3.01×1023 d-6.02 ×1023 Answer: A C19-If 58.6 g of bromine occupies 18.8 mL , then the density of bromine in g⁄mL is a-312 b-31.2 c-3.12 d-0.312 Answer: C C20-100.0 milliliter (mL) equals a-10-3 m3 b-10-4m3 c-10-5m3 d-10-6m3 Answer: B C21-The morality of hydrochloric acid (HCL) prepared by dissolving 18.25 g in 200 mL is a-1.25 M b - 1.67 M c- 2.50 M d-5.00 M Answer :C C22-What volume of a 0.05 M HCL solution is required to react with 10.0 mL of 0.02 M Na2CO3 solution according the following equation : Na2CO3 + 2 HCL→2 NaCL +CO2+H2O a-32.0 mL b-16.0 mL c-8.0 mL d-24.0 mL Answer: C C23-The number of grams of sodium hydroxide (NaOH) required to prepare a 200 mL of 0.25 M (NaOH) – solution is a-4.0 g NaOH b-3.0 g NaOH c-2.0 g NaOH d- 1.00 g NaOH Answer: C C24-How many grams of SF4 (g) can theoretically be prepared from0 . 60.0 g of SCl2 (g) and 35.0 g of NaF (s) ? The equation of reaction is 3SCl 2 (g) + 4 NaF (s) → SF4 (g) +S2Cl2 +4NaCl (s) a-21.0 g SF4 b-210 g SF4 c-2.10 g SF4 d-0.0210 g SF4 Answer: A C25-Chalcopyrite , the principal ore of copper (Cu) , contains 34.63% Cu by mass. How many grams of Cu can be obtained from 51.1 kg of the ore ? a-1.77×103 g b-1.77×104 g c-1.77×105g d-1.77×106g Answer :B C26-Which atom is most likely to from +2 ion ? a-N b-Al c-Ca d-Rb Answer: B C27-What mass of silver iodide (AgI) can be made by the reaction of 15.0 g of silver nitrate (AgNO3) with 15.0 g of sodium iodide (NaI)? a-6.91 g b- 20.74 g c-13.8 g d -27. 65 g Answer :A C28-The density of chloroform . a widely used organic solvent , is 1.4832 g⁄mL at 20°C .How many milliliters would you use if you wanted 100 g of chloroform? a-135 mL b-101 mL c-67.4 mL d-33.7 mL Answer :C C29-Iron(II) sulfate (FeSO4) is prescribed for the treatment of anemia . How many moles of FeSO4 are present in a standard 600 mg table ? a- 1.19×021 Fe2+ ions b-1.97 ×10-3 mol FaSO4 c-3.95×10-3 mol FaSO4 d-2.38×021 Fe2+ ions Answer :C C30-Acetic acid (CH3CO2H) reacts with isopentyl alcohol (C5H12O) to yield isopentyl acetate (C7H14O2) . a fragrant substance with the order of bananas .If the yield from the reaction of acetic acid with isopentyl alcohol is 65% , how many grams of isopentyl acetate are formed from 3.58 g of acetic acid and 4.75 g of isopentyl alcohol? The reaction is : CH3COOH + C5H12O→ C7H14O2 +H2O a-3.86 g b-4.56 g c-3.2 g d-2.46 g Answer: D C31-The number of moles of CO2 resulted from the reaction of 4.5 moles of C2H6 with excess oxygen according to the equation 2 C2H6 + 7O2→4 CO2 +6 H2O is: a-7.0 moles b-9.0 moles c- 10.5 moles d-13.5 moles Answer: B C32-The mass of chlorine that reacts with 2.385 g of hydrogen to from hydrogen chloride according the following equation H2 + Cl2→2HCl is: a-4.770 g b-174.1 g c-84.67 g d- 169.3 g Answer: A C33-The density of sulfuric acid in a certain car battery is 1.37 g⁄mL . what is the mass of 242 mL of the liquid? a-341 g b - 348 g c-332 g d-288 g Answer :C C34-How many grams of KOH is required to prepare a 0.5 liter of 0.35 M KOH solution ? a-5.6 g KOH b-6 g KOH c- 4.2 g KOH d-12 g KOH Answer: D C35-The molecular formula of a compound having a molecular weight 26 (g /mol) and an empirical formula CH is a-C6H6 b-C6H12 c-C2H2 d-C2H4 Answer: C C36-Genes are considered to be composed of a substance known as DNA . In particular sample , the average mass of DNA molecule is 1.0×10-16 g . Calculate the molecular mass of DNA a-6.02×108 g⁄mol b-6.02×109 g⁄mol c- 6.02×107 g⁄mol d- 6.02×1010 g⁄mol Answer: C C37-In an experiment , the actual yield was 4.35 g and the theoretical yield was 8.3 g , what is the percentage yield of the product ? a-52.4% b-64.5% c-40.4% d- 76.5% Answer: A C38-The molecular weight of KOH is a-56 g⁄mol b-63 g⁄mol c-16 g⁄mol d-40 g⁄mol Answer: A C39-The element with the number of protons=34 , neutrons=54 ,and electrons=34 is a-Ca b-Se c-K d-Au Answer: B C40- 0.90 g of sample contains 0.72 g carbon and 0.18 g hydrogen . What is the empirical formula of the compound? a-C2H3 b-CH3 c-CH2 d-CH Answer:D