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kari74 [83]
3 years ago
6

On the basis of bonding, which of the following compounds has the highest boiling point? A. water. . B. glucose. . C. sodium chl

oride. . D. alcohol
Chemistry
2 answers:
katrin2010 [14]3 years ago
5 0

Answer:

C. sodium chloride

Explanation:

Boiling point is temperature at which vapor pressure of the liquid becomes equal to external pressure (atmospheric) surrounding liquid.

On basis of bonding, the compound which has the highest boiling point is sodium chloride as it posses ionic bonding which is comprised of strong electrostatic bonds.

Water and alcohol are liquid and also, glucose has covalent bonding.

The melting points are shown below as:

Alcohol - 78.37 °C.

Glucose - 146 °C.

Water - 100°C.

Sodium chloride - 1413 °C.

<u> Answer - C. sodium chloride</u>

baherus [9]3 years ago
3 0
Alcohol = 78c
sodium chloride = 1413c
water = 100c
<span>Glucose does not have a boiling point firstly it decomposes into carbon and H2O then boils

hope it helps.
</span>


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What is a chemical property of soda ash
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A block of iron has a mass of 826 g. What is the volume of the block of iron whose density at 25°C is 7.9 ?
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Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is
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Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

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\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

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2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)

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

\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]

We are given:

\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

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Molar mass of B_5H_9 = 63.12 g/mole

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Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

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