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3 years ago

What is the physical state of a straight-chain hydrocarbon containing three carbon atoms at room temperature?

Chemistry

2 answers:

WARRIOR [948]3 years ago

8 0

Answer: Gaseous

Explanation: Straight Chain hydrocarbon having three carbon atoms is known as Propane. It molecular formula is

CH_{3}CH_{2}CH_{3}

Inter molecular forces of attraction that is London Dispersion Forces existing between its molecules are weaker that is why the propane is gaseous in nature at room temperature.

It is also used in LPG. Boiling Point of propane is -42° C.

madam [21]3 years ago

6 0

The straight-chain hydrocarbon containing three carbon atoms is propane. It's molecular formula is C3H8.

Following is the structure of propane
CH3-CH2-CH3

Boiling point of propane is -42 oC, this suggest that propane is gas at room temperature.

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arsen [322]

Answer:

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Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

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