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Amanda [17]
3 years ago
5

How is the bond in F2 different from the bond in KCl?

Chemistry
1 answer:
vagabundo [1.1K]3 years ago
6 0
It’s C because is covalent
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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

Therefore, the energy of combustion of acetylene in kJ/mol is 1097 kJ/mol.

7 0
4 years ago
Which of the following is an example of very early scientific research?
goldenfox [79]

Answer:

banging rocks together to see which make make a spark and create fire painting pictures is an example of art and sleeping in caves is survival

7 0
4 years ago
The octane rating of gasoline is a measure of the resistance to premature ignition, also known as engine knock. It is based on c
SIZIF [17.4K]

Answer: From the attached document

1. Complete structure

2. Skeletal structure

5 0
3 years ago
PLEASEEEE HELP MEEE
Nookie1986 [14]
The answer is i don’t know because i’m just doing this for points sorry!!! hope this helps
4 0
3 years ago
The volume of a 2.49 g sample of gas is 752 mL at 1.98 atm and 62 C What is the gas?
creativ13 [48]

Answer:

The volume of a sample of gas (2.49 g) was 752 mL at 1.98 atm and 62∘C 62 ∘ C .

6 0
3 years ago
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