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aleksley [76]
3 years ago
6

Find the mass of calcium oxide produced from the thermal decomposition of 50 g of calcium carbonate.

Chemistry
1 answer:
anastassius [24]3 years ago
3 0

Answer:

28 g

Explanation:

The chemical equation of the reaction is given below:

CaCO₃ (s) ⇒ CaO (s) + O₂ (g)

From the chemical equation, 1 mole of CaCO₃ produces 1 mole of CaO.

So, we find the number of moles,n of  CaCO₃ in 50 g from

n = m/M where n = number of moles of  CaCO₃, m = mass of CaCO₃ and M = molar mass of CaCO₃ = 40 g/mol + 12 g/mol + 3 × 16 g/mol = 52 g/mol + 48 g/mol = 100 g/mol

So, n = m/M

= 50 g/100 g/mol

= 0.5 mol

Since From the chemical equation, 1 mole of CaCO₃ produces 1 mole of CaO, then 0.5 × 1 mole of CaCO₃ produces 0.5 × 1 mole of CaO, which is 0.5 mole of CaCO₃ produces 0.5 mole of CaO.

So, we have 0.5 mol of CaO.

We find the mass of CaO from

n =m'/M' where n = number of moles of CaO = 0.5, m' = mass of CaO and M' = molar mass of CaO = 40 g/mol + 16 g/mol = 56 g/mol

So, m' = nM'

= 0.5 mol × 56 g/mol

= 28 g

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1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction
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H_{comb}=-4406kJ/mol

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In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

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H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol

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The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperatur
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Answer:

the activation energy Ea = 179.176 kJ/mol

it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

Explanation:

From the given information

T_1 = 100^0 C = 100+273 = 373 \ K \\ \\  T_2 = 113^0 C = 113 + 273 = 386 \ K

R_1 = \dfrac{1}{7}

R_2 = \dfrac{1}{49}

Thus; \dfrac{R_2}{R_1} = 7

Because at 113.0°C; the rate is 7 time higher than at 100°C

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In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})

1.9459 = \dfrac{Ea}{8.314}* 9.0292  *10^{-5}

1.9459*8.314 = Ea * 9.0292*10^{-5}

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Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}

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here;

T_2 = 386 \  K  \\ \\T_1 = (89.8 + 273)K = 362.8 \ K

In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})

In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})

In (\dfrac{R_2}{R_1}) = 0.00357

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\dfrac{R_2}{R_1}= 1.0035

where ;

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R_1 = \dfrac{1}{t}

Now;

\dfrac{t}{7}= 1.0035

t = 7.0245 mins

Therefore; it will take  7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.

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