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3 years ago

Rb2co3(aq)+fec2h302(aq)-->

Chemistry

1 answer:

nata0808 [166]3 years ago

4 0

Answer:

Rb2CO3(aq)+Fe(C2H3O2)2(aq)--> 2Rb(C2H3O2)(aq) + FeCO3(s)

Explanation:

The reaction shown in the answer is the reaction of rubidium carbonate and iron II acetate. Rubidium is far more reducing than Fe II hence it can displace Fe II from its salt as shown.

The reducing property of metals depends on the value of their individual electrode potential values. For rubidium, its standard reduction potential is -2.98 V while that of Fe II is -0.44V. Hence rubidium can displace Fe II from its salt as shown above.

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Kerosene, a common space-heater fuel, is a mixture of hydrocarbons whose "average" formula is C₁₂H₂₆.

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The calculated enthalpy of formation of kerosene is 365.4 kJ and heat produced is 78650.3 kJ

For this, we need the normal enthalpy of formation given below

ΔH∘f(CO2)=−393.5kJ/molΔH∘f(H2O)\s=−241.8kJ/molΔH∘f(O2)=0kJ/mol

We shall now determine the enthalpy of kerosene formation:

H rxn = 24 mol H f (CO2) + 26 mol H f (H2O) + 2 mol H f (C12H26) + 37 mol H f (O2) + 1.50 104 kJ = 9444 kJ + 6286.8 kJ + 1500 kJ 2 mol H f (C12H26) = 730.8 kJ H f (C12H26) = 365.4 kJ

Kerosene has a density of 0.74 g/mL.

Kerosene volume (V) equals 0.63 gallons, or 0.63 x 3785.4, or 2384. 8 mL.

We shall now calculate the mass (m) of kerosene:

ρ=mVm\s=ρ×Vm\s=0.749g/mL×2384.mLm\s=1786.2g

We shall now discover the heat that 1786 generated.

Two grams of kerosene

Kerosene's molar mass is 170.33 g/mol.

The mass of two moles of kerosene is equal to 2*170.33*340.66g.

1.50104kJ of heat are generated by 340.66 g of kerosene.

1786 produced heat.

Kerosene 2 grams = 1.50 104 kJ 340.66 1786.2 g = 78650.3 kJ

Learn more about enthalpy here-

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