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diamong [38]
3 years ago
10

For each reaction, write the chemical formulae of the oxidized reactants in the space provided. Write the chemical formulae of t

he reduced reactants in the space provided. 2Fe(s) 3CuCl2(aq)
Chemistry
1 answer:
Annette [7]3 years ago
5 0

Answer : 'Fe' is oxidized reactant  and 'Cu' is reduced reactant.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The balanced redox reaction will be:

2Fe(s)+3CuCl_2(aq)\rightarrow 2FeCl_3(aq)+3Cu(s)

In this reaction, the oxidation state of 'Fe' changes from (0) to (+3) that means 'Fe' lost 3 electron and it shows oxidation reaction and the oxidation state of 'Cu' changes from (+2) to (0) that means 'Cu' gain 2 electron and it shows reduction reaction.

Thus, 'Fe' is oxidized and 'Cu' is reduced.

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defon

Answer:

C1V1=C2V2

C1 is 2.0mol/l

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C2=.4mol/L

V2=100ml or for this 0.1L

V1 is 20ml

Best way to prepare this is to measure out 20ml of the 2 molar solution and add 80mL to it to get to 100mL

Explanation:

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2 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

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ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

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3 years ago
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3 years ago
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Firlakuza [10]

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