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ahrayia [7]
3 years ago
5

Walls made of paraffin wax, a covalent compound, help keep the temperature in a room steady as night changes into day and day in

to night. What is the most important property of covalent compounds that allows paraffin wax to help keep a room’s temperature level? The wall would not absorb or release any heat energy from the room or outside, so it would not affect the room’s temperature. The wall would not absorb heat during the day, but at night it would release heat that naturally exists inside of most covalent compounds. The wall would absorb extra heat during the day when the sun is out, then release the heat back into the room when the sun goes down. The wall would absorb extra heat during the day, and the wall would hold this heat in so that it cannot absorb any more heat from the night air.
Chemistry
2 answers:
Over [174]3 years ago
8 0
The wall would absorb extra heat during the day when the sun is out, then release the heat back into the room when the sun goes down.
MAXImum [283]3 years ago
7 0

Answer: Option (c) is the correct answer.

Explanation:

Covalent compounds are the compounds in which atoms combine by sharing of electrons.

These compounds have weak forces of attraction and therefore, they easily heat up and thus, have low melting and boiling point.

Hence, the most important property of covalent compounds that allows paraffin wax to help keep a room’s temperature level is that the wall would absorb extra heat during the day when the sun is out, then release the heat back into the room when the sun goes down.

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Which of the following elements has six valence electrons?<br> Be<br> B<br> C<br> N<br> O
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2 years ago
At 298 K, the rate constant for a reaction is 0.0346 s-1. What is the rate constant at 350K if the Ea = 50.2kJ/mol
frutty [35]

Answer:

0.702 /s

Explanation:

Rate constant at [298 \mathrm{~K}, \mathrm{~K}_{1}=3.46 \times 10^{-2} \mathrm{~s}^{-1}

Rate constant at 350 \mathrm{~K}, \mathrm{~K}_{2}=?

T_{1}=298 \mathrm{~K}

T_{2}=350 \mathrm{~K}

Activation energy, \mathrm{Ea}=50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}

Use the following equation to calculate K_{2}$ at $350 \mathrm{~K}

Use the following equation to calculate K_{2}$ at $350 \mathrm{~K}

\ln \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{\mathrm{R}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{\mathrm{~T}_{2}}\right]

Therefore,

 \ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{1}{298 \mathrm{~K}}-\frac{1}{350 \mathrm{~K}}\right]

\ln \left(\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}}\right) &=\frac{50.2 \times 10^{3} \mathrm{~J} / \mathrm{mol}}{8.314 \mathrm{JK}^{-1} \mathrm{~mole}^{-1}}\left[\frac{52 \mathrm{~K}}{298 \mathrm{~K} \times 350 \mathrm{~K}}\right]

\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=\mathrm{e}^{3.01}

\frac{K_{2}}{3.46 \times 10^{-2} \mathrm{~s}^{-1}} &=20.3

K_{2} &=20.3 \times 3.46 \times 10^{-2} \mathrm{~s}^{-1}

&=0.702 \mathrm{~s}^{-1}

hence, the rate constant at 350 \mathrm{~K} is 0.702\mathrm{~s}^{-1}

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