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3 years ago

Which of the following is a chemical property? A. Boiling point B. Electrical conductivity C. Flammability D. Malleability

Chemistry

1 answer:

Scrat [10]3 years ago

7 0

Answer:

I say its C flammability

Explanation:

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In the laboratory a general chemistry student finds that when 2.84 g of KClO4(s) are dissolved in 107.70 g of water, the tempera

vredina [299]

Answer : The enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $1.55J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 107.70 g

$\Delta T$ = change in temperature = $T_2-T_1=(22.80-20.34)=2.46^oC$

Now put all the given values in the above formula, we get:

$q=[(1.55J/^oC\times 2.46^oC)+(107.70g\times 4.184J/g^oC\times 2.46^oC)]$

$q=1112.3J=1.1123kJ$

Now we have to calculate the enthalpy change of dissolution of $KClO_4$

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 1.1123 kJ

m = mass of $KClO_4$ = 2.84 g

Molar mass of $KClO_4$ = 138.55 g/mol

$\text{Moles of }KClO_4=\frac{\text{Mass of }KClO_4}{\text{Molar mass of }KClO_4}=\frac{2.84g}{138.55g/mole}=0.0205mole$

$\Delta H=\frac{1.1123kJ}{0.0205mole}=54.3kJ/mole$

Therefore, the enthalpy change of dissolution of $KClO_4$ is 54.3 kJ/mole

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4 years ago

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Norma-Jean [14]

Gaseous matter has no definite shape because the particles are packed so loosely and are very energetic, moving a lot.

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4 years ago

An atom of argon rarely bonds to an atom of another element because an argon atom has

Solnce55 [7]

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If a wave produced by a fireworks explosion is that light wave or sound wave

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The wave produced by a fireworks "explosion" is a light wave.

Explanation:

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