Question

Section A: (Circle or bubble the correct answer with PEN) Which solution of a base contains the smallest number of moles of a base? a) 5.0 L of 4.0 M b) 0.50 L of 4.0 M c) 4.0 L of 0.40 M d) 0.50 L of 0.40 M What is the volume of 0.1220 M HNO3 solution required to completely neutralize 17.50 mL of 0.1211 M NaOH ? 3 Reaction is: HNO; (aq) + NaOH(aq) → NaNO3(aq) + H20 (aq) a) 17.84 mL b) 16,31 mL c) 15.39 mL d) 17.37 mL Using the following portion of the activity series of elements Mg(8) ► Mg2+ (aq) + 2e- (strongly reducing) Al(s) > Al+(aq) + 3e- Cu(s) → Cu²+(aq) + 2e- (weakly reducing) Determine which one of the reactions will occur. a) Cu(s) + Ca2+(aq) b) Al(s) + Ca 2+ (aq) 4 b) Al(s) + Cu(aq) d) Cu(s) + Al3+ (aq) In the following combustion reaction of methane gas (CH4) with oxygen is given. How many (g) of H20 is produced when 0.533 moles of CH4 reacts with O2? Reaction is: CH4 + 202 → CO2 + 2H30 a) 19.21 g b) 17.21 g c) 9.60 d) 15.23 g 5 How many grams (g) of NaCl are needed to prepare 250.0 ml of 0.365 M of NaCl. Solution? a) 3.65 g b) 10 25 g c) 5.34 g d) 7.21 g If 0.500 L of a 3.50 M HCl acid is diluted to 1.00 L, the molarity HCl acid is diluted to 1.00 L, the molarity of the new solution is: a) 1.50 M b) 3.50 M c)1.40 M d) 1.75 M

Answer 1

Naotlcag + HINO cag - Nanozcaq) + H2oce) Poup Calculale No. of moles of Naat Moles = Mass Molar Mass of NaOH = 39.997 glmof Molar Mass Given Molarity of HNO3 = 0.1220M Volume of HNO3 = ? Volume of NaoH= 17. 5mL Molarity of NaOH = 0.1211 Molarity = no. of moles xwoo VeinmL) Molarity = Max OOO Molar mas XVML) mass = Molarity x Molas mans & vin me) 1000 For NAOH man s 0.1221817.5x39.997 = 0.0847 1000 : No. of moles of Naot = 0.0847 39.997 =0.002117 = 2.117 mmoll Mole Ra 03: NaoH = 1:1 Therefore if moles of MaoH is 2.117 mmol, then HNO3 is also 2.117 mmol Molanty - moles x 1000 VinmL) Ven mi) = 2.117 mmol x1000 = 17.37m4 Option (d) is correct

We have , Molarity - no. of moles. Vem) No. of moles - Molarity & Vane). No of moles & Volume. lowest smallest The one with lowest volume will have number of moes at a constant molarity. Here there are 4 cases. a seraf notes a) 5-OL of H-OM No. of moles = Molanity Volume co = 5x4=20moles (b) 0.5L of 4.0m 0.5* 4= No. of molus = 2 mokes (C) 4.0L of 0.40M No. of moles = LOK 0.4 = 16moles (6) 0.50L of 040m les = 0. 50 80.40=0.2 moles There option ( 0.50L of 0.40m how smallest neember of moles.

3.Activity Series of Metals

Metal	Symbol	Reactivity
Lithium	Li	displaces H2 gas from water, steam and acids and forms hydroxides
Potassium	K
Strontium	Sr
Calcium	Ca
Sodium	Na
Magnesium	Mg	displaces H2 gas from steam and acids and forms hydroxides
Aluminum	Al
Zinc	Zn
Chromium	Cr
Iron	Fe	displaces H2 gas from acids only and forms hydroxides
Cadmium	Cd
Cobalt	Co
Nickel	Ni
Tin	Sn
Lead	Pb
Hydrogen gas	H2	included for comparison
Antimony	Sb	combines with O2 to form oxides and cannot displace H2
Arsenic	As
Bismuth	Bi
Copper	Cu
Mercury	Hg	found free in nature, oxides decompose with heating
Silver	Ag
Palladium	Pd
Platinum	Pt
Gold	Au

The top metals are more reactive than the metals on the bottom. So the metals on the top can undergo displacement reaction with the metal present bottom to it.

(a) This reaction does not occur as Ca is placed above Cu.Ca is more reactive than Cu and no reaction takes place

(b)This reaction does not occur as Ca is placed above Al.Ca is more reactive than Al and no reaction takes place

(C)This reaction does occur as Al is placed above Cu. Al is more reactive than Cu and hence  reaction takes place

(d)This reaction does not occur as Al is placed above Cu. Al is more reactive than Cu and no reaction takes place

Poup CH4 + 202 CO2 + 2H2o. Imole of CH4 a moles of wales 0533 moles of CH4 - 2x0.533 moles of H2O - 1066moles of 4120 کر انکار کا کردار morar mais = mass= 1.0664 18,01528 ginol = 19.219 Option is correct 841 Molarity = moss 1000 molaamass & Venu) mon = Molarity x molas manx Vime) looo 0.365mx250x58.44 g/mol tooo 5.349 Option (C) is correct 6 MIVI= M2 42 Dilution Formula Mi= ao 3.5m M2=2 MIVI = M2V2 M2 = MIVI 3.500.5 Vi= 0.5L V2 V2= 1 option (d) = 1:15M