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klemol [59]
3 years ago
12

What are three examples of fields?

Chemistry
1 answer:
Shalnov [3]3 years ago
5 0
Answer:

Here are all the fields in Chemistry. I hope this answers your question.

Explanation:


There are five main types of chemistry: organic, inorganic, physical, analytical, and biochemistry.
You might be interested in
Classify each chemical reaction:
I am Lyosha [343]

Answer:

1) combustion

2) double replacement

3) combination

4) combustion

Explanation:

The combustion of a compound refers to the reaction of that compound with oxygen to produce heat and light. In reactions (1) and (4) above, ethanol and methane reacted with oxygen to yield carbon dioxde and water. This is a combustion reaction.

Reaction(2) is a double replacement reaction because the both cations exchange their anion partners in the product.

Reaction (3) is a combination reaction. It involves the joining of two elements to form a new compound.

5 0
3 years ago
. 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
Cylinder of air at 1.5 atm of pressure is kept at room temperature while a piston compresses the air from 40 l down to 10 ml. wh
djyliett [7]

The new pressure, P₂ is 6000 atm.

<h3>Calculation:</h3>

Given,

P₁ = 1.5 atm

V₁ = 40 L = 40,000 mL

V₂ = 10 mL

To calculate,

P₂ =?

Boyle's law is applied here.

According to Boyle's law, at constant temperature, a gas's volume changes inversely with applied pressure.

                                        PV = constant

Therefore,

P₁V₁ = P₂V₂

Put the above values in the equation,

1.5 × 40,000 = P₂ × 10

P₂  = 1.5 × 4000

P₂  = 6000 atm

Therefore, the new pressure, P₂ is 6000 atm.

Learn more about Boyle's law here:

brainly.com/question/23715689

#SPJ4

8 0
2 years ago
Suppose you need to make the following two precipitates (by two different precipitation reactions): MgCO3 and Ca3(PO4)2. For eac
Setler [38]

Answer:

For the first reaction the reagents are: MgCl2 and Na2CO3

For the second reaction the reagents are: Na2HPO4 and CaCl2

Explanation:

Precipitation reactions lie in the production of a compound that is not soluble, which is called a precipitate, this precipitate is produced when two different solutions are combined, each of which will contribute an ion for the formation of the precipitate. In the first reaction you have:

MgCl2 + Na2CO3 = MgCO3 + 2 NaCl

Type of reaction: double displacement

The second reaction is as follows:

4Na2HPO4 + 3CaCl2 → Ca3 (PO4) 2 + 2NaH2PO4 + 6NaCl

It is the reaction of sodium hydrochlorophosphate and calcium chloride

3 0
3 years ago
How many molecules are there in 223 grams of Na2SO4?
Harlamova29_29 [7]
<h3><u>Answer;</u></h3>

= 9.45 × 10^23 molecules

<h3><u>Explanation; </u></h3>

The molar mass of Na2SO4 = 142.04 g/mol

Number of moles = mass/molar mass

                             = 223/142.04

                             =  1.57 moles

But;

1 mole = 6.02 × 10^23 molecules

Therefore;

1.57 moles = ?

= 1.57 × 6.02 × 10^23 molecules

<u>= 9.45 × 10^23 molecules </u>

8 0
3 years ago
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