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Sladkaya [172]
3 years ago
13

Consider the balanced chemical reaction below. When the reaction was carried out, the calculated theoretical yield for carbon di

oxide was 93.7 grams, but the measured yield was 88.3 grams. What is the percent yield?
Fe2O3 + 3CO --> 2Fe + 3CO2
Chemistry
2 answers:
bekas [8.4K]3 years ago
8 0

Answer:94.23

Explanation:

Hope it helps

Arada [10]3 years ago
3 0

The percent yield represents that the efficiency of a reaction.

It represents how much times we have achieved the target yield from a particular reaction

The relation between theoretical yield, measured yield and percent yield is

percent yield = \frac{measuredyieldX100}{theoreticalyield}

for the given reaction

percent yield = \frac{88.3X100}{93.7}=94.23%

Answer: 94.23%

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calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
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Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

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Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

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T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

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[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

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