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Inessa [10]
4 years ago
11

Sodium sulfate reacts with carbon to produce the products sodium sulfide and carbon monoxide. Identify the reducing agent in the

following reaction.
Na2SO4 + 4C yields Na2S + 4CO

Na2SO4
C
Na2S
CO
Chemistry
2 answers:
Lunna [17]4 years ago
7 0

Answer : The correct option is, carbon (C)

Explanation :

Oxidizing agent : Oxidizing agent are the agent that gains electrons and is reduced in a chemical reaction.

Reducing agent : Reducing agent are the agent that lose electrons and is oxidized in a chemical reaction.

The given balanced chemical reaction is,

Na_2SO_4+4C\rightarrow Na_2S+4CO

First we have to calculate the oxidation number of sulfur and carbon.

The oxidation state of 'S' in Na_2SO_4 and Na_2S are, (+6) and (-2) respectively.

The oxidation state of 'C' in 4C and 4CO are, (0) and (+2) respectively.

From this we conclude that, the Na_2SO_4 act as an oxidizing agent and 'C' act as a reducing agent.

Hence, the reducing agent in the following reaction is, carbon (C)

Maslowich4 years ago
6 0
The reducing agent will itself be oxidized.
The oxidation number of carbon goes form 0 to +2. Therefore, it is the reducing agent.
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A) Calculate the osmotic pressure difference between seawater and fresh water. For simplicity, assume thatall the dissolved salt
never [62]

Answer:

a)  Δπ = 1.264 atm

b) W = 128 joules

c)  ΔH >> W  ( a factor greater than 17,000 )

Explanation:

a) The osmotic pressure, π , is determined by :

π = nRT/V, where n= moles of solute

                          R= 0.0821 Latm/kmol

                          T = 300 K

calling π(sw) osmotic pressure for  for sea water and π (fw) for fresh water,

salinity of sea water = 3.5 g / 1L water   (assuming only NaCl for the salts)

salinity of fresh water = 0.5 parts per thousand (range: 0- 0.5 ppt)

πsw = (3.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 1.475 atm

πfw = (0.5 g/58.44 g/mol) (0.0821 Latm/Kmol) (300 K ) /1 L = 0.211 atm

d water = 1 g/cm³

Δ π = (1.475 - 0.211) = 1.264 atm

b) W = Δπ V = 1.426 atm x 1L = 1.43 L-atm

1 L-atm = 101.33 j

W =  101.33 j/ Latm x  1.43 Latm = 128 joules

c) ΔH = Q₁ + nΔH vap, where

            Q₁  = heat required to bring the solution from 300 K to boiling, 373 K

            ΔH vap = heat of vaporization

Q = mCΔT = 1000 g x 4.186 j x 73 K = 305.6 j = 0.3056 kj

ΔH vap = (1000 g/ 18 g/mol ) 40.7 kj/mol = 2,261 kj

ΔH =  0.3056 kj + 2,261 kj = 2,261.3 kj

Note = Q << ΔH vap and we could have neglected it.

This result shows why nobody talks about evaporation of sea water to produce fresh water ΔH >> W

6 0
3 years ago
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