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klio [65]
3 years ago
5

Classify the reaction between zinc and acetic acid. Please answer fast, due tomorrow!

Chemistry
2 answers:
LenKa [72]3 years ago
7 0
1) Zn(s) + 2 HO2CCH3(l) --> Zn(O2CCH3)2 + H2(g)

There are two half reactions:
Zn -->Zn (2+) + 2e-
2e- + 2 H+ --> H2

(the acetic acid ionized to form H+ and acetate- ions)


2) It's a red-ox (reduction-oxidation) reaction. The Zn gives up 2 electrons to the 2 hydrogen atoms. In general one element loses electrons and another gains electrons. The element gaining the electrons is the oxidizer and the element losing them is the reducing agent. The oxidizer becomes reduced and the reducing agent becomes oxidized.

3) Hydrocarbons and oxygen.
Bess [88]3 years ago
5 0
Zn + 2HCl → ZnCl2 + H2 This reaction is rapid, because HCl is a strong acid and provides more reactive H+ ions. The products are zinc chloride and hydrogen gas. Zn + 2CH3COOH → Zn(CH3COO)2 + H2 This reaction is slower, because acetic acid is a weak acid. The products are zinc acetate and hydrogen gas.

So I would classify it as expeditious
Hope it helps!
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3 0
2 years ago
In a coffee-cup calorimeter, 150.0 mL of 0.50 M HCl is added to 50.0 mL of 1.00 M NaOH to make 200.0 g solution at an initial te
Zigmanuir [339]

Answer:

51.54°C the final temperature of the calorimeter contents.

Explanation:

HCl+NaOH\rightarrow H_2O+NaCl,\Delta H=-56 kJ/mol

moles=Molarity\times Volume (L)

Molarity of HCl= 0.50 M

Volume of HCl= 150.0 mL = 0.150 L

Moles of HCl= n

n=0.50 M\times 0.150 L=0.075 mol

Molarity of NaOH= 1.00 M

Volume of NaOH= 50.0 mL = 0.050 L

Moles of NaOH= n'

n'=1.00 M\times 0.050 L=0.050 mol

Since moles of NaOH are less than than moles of HCl. so energy release will be for neutralization of 0.050 moles of naOH by 0.050 moles of HCl.

n = 0.050

-56 kJ/mol=-\frac{Q}{n}

Q=56 kJ/mol\times 0.050 kJ/mol=2.8 kJ=2800 J

(1 kJ= 1000 J)

The energy change released during the reaction = 2800 J

Volume of solution = 150.0 mL + 50.0 mL = 200.0 mL

Density of the solution (water) = 1.00g/mL

Mass of the solution , m= 200 mL × 1.00 g/mL = 200 g

Now , calculate the final temperature by the solution from :

q=mc\times (T_{final}-T_{initial})

where,

q = heat gained = 2800 J

c = specific heat of solution = 4.184 J/^oC

T_{final} = final temperature = ?

T_{initial} = initial temperature = 48.2^oC

Now put all the given values in the above formula, we get:

2800 J=200.0 g\times 4.184 J/^oC\times (T_{final}-48.2)^oC

T_{final}= 51.54^oC

51.54°C the final temperature of the calorimeter contents.

7 0
3 years ago
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