Question

can you please do 11,12,and 13

8 The Following teaction is elassified as: T 3HT T0 333LIO 283 D 2HCI (aq)+ Ba(OH)p(ag) BaCh (aq)+2 Hy0) d lo C00-10 M a) oxidation-reduction reaction b) precipitation reaction e) single replacement d) acid-base neutralization reaction What is the volume of 0.1220 M HNO, solution required to completely neutralize 17.50 ml. of 0.1211 M NaOH? 9. ors NaNO, (aq) +HO (aq) e a) 17.37 ml. b) 16.31 mL c) 15.39 ml. d) 17.84 m Which one of the following reactions is the reduction reaction? A lo 10. a) Cu" (aq)+ 2e Cu (s) Cr(ag)+3e b) Cris) c) Cu'Cu + e d) Fe Fe+2e 11. Using the following portion of the activity series of elements Cals)-Ca(aq) +2er (strongly reducing) Als)Al(aq) +3e Col)Cut(aq) + 2e (weakly reducing) Determine which one of the reactions will occur a) Cus)+ Ca(ag) b) Al)+Cut(a) c) Al)+ Ca(aq d) Culs) +Alt(aq) In the following combustion reaction of methane gas (CHa) with oxygen is given. How many (g) of H20 is produced when 0.533 moles of CHareacts with O27 12. a) 15.23 g b) 9.60 c) 19.21 g d) 17.21 g How many grams (g) of NaCl are needed to prepare 250.0 ml of 0.365 M of NaCL Solution? a) on b) 10 25 g e) 3.65 g d) 7.21 g 5.34 g

Answer 1

Q11. Correct option is : (b) Al(s) + Cu²⁺(aq)

Q12. Correct option is : (c) 19.21 g

Q13. Correct option is : (a) 5.34 g

Explanation

Q12. In the activity series of metals, a solid metal can reduce the cation of the metal below.

For example in the given series, Ca (s) can reduce Al³⁺ (aq) and Cu²⁺ (aq)

Similarly, Al (s) can reduce Cu²⁺ (aq)

Q13. The balanced equation is : CH₄ + 2 O₂ $\rightarrow$ CO₂ + 2 H₂O

moles CH₄ reacted = 0.533 mol

moles H₂O formed = (moles CH₄ reacted) * (2 moles H₂O / 1 mole CH₄)

moles H₂O formed = (0.533 mol) * (2/1)

moles H₂O formed = (0.533 mol) * 2

moles H₂O formed = 1.066 mol

mass H₂O formed = (moles H₂O formed) * (molar mass H₂O)

mass H₂O formed = (1.066 mol) * (18.0 g/mol)

mass H₂O formed = 19.21 g

Q13. concentration NaCl = 0.365 M

volume NaCl solution = 250.0 mL = 0.2500 L

moles NaCl required = (concentration NaCl) * (volume NaCl solution)

moles NaCl required = (0.365 M) * (0.2500 L)

moles NaCl required = 0.09125 mol

mass NaCl required = (moles NaCl required) * (molar mass NaCl)

mass NaCl required = (0.09125 mol) * (58.44 g/mol)

mass NaCl required = 5.34 g