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adell [148]
3 years ago
10

Which formula equation shows a reversible reaction?

Chemistry
2 answers:
KatRina [158]3 years ago
5 0

Its c I just took the test

allochka39001 [22]3 years ago
3 0

The formula equation that  shows   a  reversible  reaction  is

NH4Cl (s) ⇔ NH3 (g)  + HCl (g) ( answer  C)

<em>            </em>

<em>            </em><u><em> Explanation</em></u>

  • Reversible  reaction  is  a chemical reaction    where the reactants form products , where in turn react together to form reactant   back.
  • The  reaction of  NH4Cl (s)⇔ NH3(g)  + HCl (g)  is a reversible reaction  since   NH4Cl  decompose to form NH3  and HCl which in  turn react together to form NH4Cl back .
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A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by th
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Answer:

The ΔH of the reaction is + 12.45 KJ/mol

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Mass of water= 100ml = 100g. (You should always assume 1cm3 of water as 1g)

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Step 1 : Calculate the heat energy (Q) lost by the water.

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            Q = -100 x 4.18 x (-5.14)

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Step 2: Calculating the ΔH of the reaction?

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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
kipiarov [429]

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

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

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

ΔH → enthalpy of transition

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}

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(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\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

6 0
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