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2 years ago

What mass of lead (density 11.4 g/cm3) would have an identical volume to 28.5 g of mercury (density 13.6 g/cm3)?

1 answer:

6 0

The density of a material is its unique property which is different for different substances. The volume or mass of a compound can be determined from the other given mass of a compound, if the density of both the material is available. We know density of a material is equal to the mass per unit volume. From the given data 13.6 g of mercury has 1 cm³ volume. Thus 28.5g of mercury will have $\frac{28.5}{13.6}=2.095$ cm³ volume. Now as per the given density value of lead 11.4g of lead has 1 cm³ volume. Thus 2.095 cm³ volume of lead has (2.095×11.4)=23.889g.

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2 years ago

Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

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Answer:

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Explanation:

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