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Olin [163]
3 years ago
15

(ii) Arrange the following metals in decreasing order of their reactivity:Fe, Zn, Na, Cu, Ag​

Chemistry
1 answer:
ratelena [41]3 years ago
4 0

Answer:

Given Elements are

  • Ferrus(Fe)
  • Zinc(Zinc)
  • Sodium (Na)
  • Copper (Cu)
  • Silver(Ag)
<h3>Decreasing Order:-</h3>

Na>Zn>Fe>Cu>Ag

<h3><u>Expla</u><u>nation</u><u>:</u><u>-</u></h3>
  • Here is the reactivity series of elements to let you understand better .

\setlength{\unitlength}{0.8 cm}\begin{picture}(20,15)\thicklines\put(5,9.5){\large {\boxed{ \sf{Reactivity \:  series}}}}\put(0.8,8){\large\sf K \qquad\qquad\qquad Potassium\qquad\qquad\qquad \: maximum}\put(0.8,7.2){\large\sf Na \qquad\qquad\qquad Sodium\qquad\qquad\qquad\quad \: reactivity}\put(0.8,6.2){\large\sf Ca\qquad\qquad\qquad \: Calcium}\put(0.8,5.2){\large\sf Mg\qquad\qquad\quad \: Magnesium}\put(0.8,4.2){\large\sf Al\qquad\qquad\qquad \: Aluminium\qquad\qquad\qquad \: decreasing}\put(0.8,3.2){\large\sf Zn\qquad\qquad\quad\qquad \: Zinc\qquad\qquad\qquad\qquad \:reactivity}\put(0.8,2.2){\large\sf Fe\qquad\qquad\quad\qquad Ferrous}\put(0.8,1.2){\large\sf Pb\qquad\qquad\qquad \: \:\: Lead}\put(0.8,0.2){\large\sf H\qquad\qquad\qquad \:  \:Hydrogen}\put(0.8, - 0.6){\large\sf Cu\qquad\qquad\qquad \: Copper}\put(0.8, - 1.5){\large\sf Hg\qquad\qquad\qquad \: Mercury}\put(0.8, - 2.5){\large\sf Ag\qquad\qquad\qquad \: \:Silver}\put(0.8, - 3.5){\large\sf Au\qquad\qquad\qquad \: \:  Gold\qquad\qquad\qquad\quad \:  \: minimum reactivity}\put(9.3,8.2){\vector(0, - 2){12}}\end{picture}

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A forklift applies a force of 2,000 N to raise a box 3 m.
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2 years ago
Determine the pH of (a) a 0.40 M CH3CO2H solution, (b) a solution that is 0.40 M CH3CO2H and 0.20 M NaCH3CO2
elena-s [515]

Answer:

a) pH = 2.573

b) pH = 4.347

Explanation:

a) weak acid: CH3COOH

  • CH3COOH + H2O ↔ CH3COO- + H3O+

∴ Ka = 1.8 E-5 = [H3O+][CH3COO-] / [CH3COOH]

∴ <em>C</em> CH3COOH = 0.40 M

mass balance:

⇒ 0.40 M = [CH3COO-] + [CH3COOH].........(1)

charge balance:

⇒ [H3O+] = [CH3COO-].........(2)

(2) in (1):

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ Ka = 1.8 E-5 = [H3O+]² / ( 0.40 - [H3O+] )

⇒ [H3O+]² = 7.2 E-6 - 1.8 E-5[H3O+]

⇒ [H3O+]² + 1.8 E-5[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 2.6743 E-3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.573

b) balanced reations:

  • CH3COONa + H2O → Na+  +  CH3COO-
  • CH3COOH + H2O ↔ CH3COO-  +  H3O+

∴ <em>C</em> CH3COOH = 0.40 M

∴ <em>C</em> CH3COONa = 0.20 M

mass balanced:

⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [CH3COO-] + [CH3COOH]

⇒ 0.60 = [CH3COO-] + [CH3COOH]......(1)

charge balanced:

⇒ [H3O+] + [Na+] = [CH3COO-]

∴ [Na+] = 0.20 M

⇒ [H3O+] + 0.20 M = [CH3COO-]........(2)

(2) in (1):

⇒ 0.60 M = ( [H3O+] + 0.20 ) + [CH3COOH]

⇒ [CH3COOH] = 0.40 - [H3O+]

replacing in Ka:

⇒ 1.8 E-5 = ([H3O+])([H3O+] + 0.20) / (0.40 - [H3O+])

⇒ 7.2 E-6  - 1.8 E-5[H3O+] = [H3O+]² + 0.20[H3O+]

⇒ [H3O+]² + 0.20[H3O+] - 7.2 E-6 = 0

⇒ [H3O+] = 4.499 E-5 M

⇒ pH = 4.347

7 0
3 years ago
What’s the answer? I need an answer as soon as possible please!
inysia [295]

Answer:

1, 1, 2

Explanation:

1, 1, 2

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