Why do covalent compounds tend to be squishy?

Chemistry

1 answer:

Sav [38]3 years ago

8 0

Answer:

Since they're easy to separate, covalent compounds have low melting and boiling points. 2) Covalent compounds are soft and squishy (compared to ionic compounds, anyway). The reason for this is similar to the reason that covalent compounds have low melting and boiling points. When you hit an ionic compound with something, it feels very hard

Explanation:

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. The heat capacity of a bomb calorimeter was determined by burning 6.79 g methane (energy of combustion 802 kJ/mol CH4) in the

kolbaska11 [484]

Answer:

a) The heat capacity of the calorimeter is 31.4 kJ/ºC.

b) The energy of combustion of acetylene in kJ/mol is 1097 kJ/mol.

Explanation:

The heat capacity ( C ) of a substance is <em>the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius</em>. Its units are J/°C. or kJ/ºC.

If we know the heat capacity and the amount of a substance, then the change in the sample’s temperature (Δt ) will tell us the amount of heat (<em>q</em>) that has been absorbed or released in a particular process. One of the equations for calculating the heat change is given by:

$q=C.ΔT$

Where ΔT is the temperature change: ΔT= tfinal - tinitial, and C the heat capacity.

In the calorimeter, the heat given off by the sample is absorbed by the water and the bomb. The special design of the calorimeter enables us to assume that no heat (or mass) is lost to the surroundings during the time it takes to make measurements.

Therefore, we can call the bomb and the water in which it is submerged an isolated system. Because no heat enters or leaves the system throughout the process, the heat change of the system ( q system ) must be zero and we can write:

$qsystem = qrxn + qcal$

$qsystem = 0$

where q cal and q rxn are the heat changes for the calorimeter and the reaction, respectively. Thus, $qrxn = -qcal$

To calculate <em>q</em>cal , we need to know the heat capacity of the calorimeter ( Ccal ) and the temperature rise, that is, <em> $qcal = Ccal. ΔT$ </em>

a. The quantity Ccal is calibrated by burning a substance with an accurately known heat of combustion. In order to do this, we need to convert the molar heat of combustion (expressed in kJ/mol) into heat of combustion (expressed in kJ). For that matter, we transform the 6.79 grams of methane into moles:

$1 mol CH₄÷16.04 g CH₄ × 6.79 g CH₄ = 0.423 mol CH₄$

And then multiply it by the molar heat of combustion:

$802 kJ/mol × 0.423 mol = 339 kJ$

Now we know that that the combustion of 6.79 g of methane releases 339 kJ of heat. If the temperature rise is 10.8ºC, then the heat capacity of the calorimeter is given by

$Ccal= qcal/ΔT = 339 kJ/10.8ºC = 31.4 kJ/ºC$

Once C cal has been determined, the calorimeter can be used to measure the heat of combustion of other substances. Note that although the combustion reaction is exothermic, q cal is a positive quantity because it represents the heat absorbed by the calorimeter.

b. The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with the same equation, and assuming no heat is lost to the surroundings, we write

$qcal=Ccal.ΔT= 31.4 kJ/°C × 16.9 °C = 531kJ$

Now that we have the heat of combustion, we need to calculate the molar heat. Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -531 kJ.

This is the heat released by the combustion of 12.6 g of acetylene ; therefore, we can write the conversion factor as 531 kJ÷12.6 g

The molar mass of acetylene is 26.04 g, so the heat of combustion of 1 mole of acetylene is

$molar heat of combustion= -531 kJ÷12.6 g × 26.04 g÷ 1 mol= 1097 kJ/mol$