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3 years ago

The vaporization of Br2 from the liquid gas to the solid state requires 7.4 kcal/nil what is the triangleH for this process

1 answer:

5 0

The initial state of the compound Br2 is liquid. The final state is then solid. We know for certain that the liquid state has higher energy as compared to solid state. Due to this fact, Br2 must release energy to be in solid form. Therefore, the change in enthalpy is:

ΔH = -7.4 kcal/nil

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A hot air balloon filled with 1.27x10^6L of an ideal gas on a cool morning of 11 degrees C. the air is heated to 105 degrees C.

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Answer:

New volume = $1.69\times 10^6$ L

Explanation:

Using Charle's law

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

V₁ = $1.27\times 10^6$ L

V₂ = ?

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The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (11 + 273.15) K = 284.15 K

T₂ = (105 + 273.15) K = 378.15 K

Using above equation as:

$\frac{1.27\times 10^6}{284.15}=\frac{V_2}{378.15}$

$V_2=\frac{1.27\times 10^6\cdot \:378.15}{284.15}$

<u>New volume = $1.69\times 10^6$ L</u>

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