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ANEK [815]
3 years ago
5

+ c) FeCl3 + NH4OH Fe(OH)3 NHACI

Chemistry
1 answer:
bixtya [17]3 years ago
5 0

The question is incomplete, the complete question is:

Write the net ionic equation for the below chemical reaction:

(c): FeCl_3+3NH_4OH\rightarrow Fe(OH)_3+3NH_4CI

<u>Answer:</u> The net ionic equation is Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)

<u>Explanation:</u>

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

(c):

The balanced molecular equation is:

FeCl_3(aq)+3NH_4OH(aq)\rightarrow Fe(OH)_3(s)+3NH_4Cl(aq)

The complete ionic equation follows:

Fe^{3+}(aq)+3Cl^-(aq)+3NH_4^+(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)+3NH_4^+(aq)+3Cl^-(aq)

As ammonium and chloride ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

Fe^{3+}(aq)+3OH^{-}(aq)\rightarrow Fe(OH)_3(s)

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A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C
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Explanation:

The given data is as follows.

     V_{1} = 50 ml,      T_{1} = 345 K

     T_{2} = 298 K,      T_{f} = 317 K,

    V_{2} = 50 ml

Now, we will calculate the heat energy as follows.

        Q_{hot} = m \times C_{p} \times (T_{1} - T_{f})

                     = 50 g \times 4.184 \times (345 - 317)

                     = 5857.6 J

       Q_{cold} = m \times C_{p} \times (T_{f} - T_{2})

                     = 50 g \times 4.184 \times (317 - 298)

                     = 3974.8 J

As,   Q_{hot} = -Q_{cold} so there will be loss of heat. And, some heat will go to the calorimeter.

Hence,     Q_{hot} = Q_{cold} + Q_{cal}

                 5857.6 = 3974.8 + Q_{cal}

               Q_{cal} = 1882.8 J

We assume that the temperature of (calorimeter + water) is 298 K. Hence,

                  dT = T_{f} - T_{2}

                        = (317 - 298) K

                        = 19 K

Hence, we will calculate the specific heat as follows.

               C = \frac{Q}{dT}

                   = \frac{1882.8 J}{19}

                   = 99.1 J/K

Thus, we can conclude that the value of C_{cal} for the calorimeter is 99.1 J/K.

3 0
3 years ago
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