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2 years ago

Explain the three types of cumbusion ??

Chemistry

1 answer:

Alik [6]2 years ago

6 0

Answer:

Complete Combustion

One of the types of combustion is Complete Combustion. Complete combustion occurs in an unlimited supply of air, oxygen in particular. Also, complete combustion is also known as clean combustion. Here the hydrocarbon will burn out completely with the oxygen and leave only two byproducts, water, and carbon dioxide.

An example of this is when a candle burns. The heat from the wick will vaporize the wax which reacts with the oxygen in the air. The two products of the reaction are water and carbon dioxide. In an ideal situation all the wax burns up and complete combustion takes place

2] Incomplete Combustion

Incomplete combustion takes place when the air is in limited supply. And as opposed to complete combustion it is otherwise known as dirty combustion. Due to lack of oxygen, the fuel will not react completely. This, in turn, produces carbon monoxide and soot instead of carbon dioxide.

An example is burning of paper. It leaves behind ash (a form of soot) as a byproduct. In a complete combustion, the only products are water and carbon dioxide. Also, incomplete combustion produces less energy than complete combustion.

3] Rapid Combustion

Another type of combustion is Rapid Combustion. Rapid energy needs external heat energy for the reaction to occur. The combustion produces a large amount of heat and light energy and does so rapidly. The combustion will carry on as long as the fuel is available.

An example is when you light a candle. The heat energy is provided when we light the candle with a matchstick. And it will carry on till the wax burns out. Hence it is a rapid combustion

4] Spontaneous Combustion

As the name suggests the combustion occurs spontaneously. This means that it requires no external energy for the combustion to start. It happens due to self-heating. A substance with low-ignition temperatures gets heated and this heat is unable to escape.

The temperature rises above ignition point and in the presence of sufficient oxygen combustion will happen. The reaction of alkali metals with water is an example.

5] Explosive Combustion

Explosive Combustion happens when the reaction occurs very rapidly. The reaction occurs when something ignites to produce heat, light and sound energy, The simple way to describe is it to call it an explosion. Some classic examples are firecrackers or blowing up of dynamite.

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Write a chemical equation for the following decomposition reaction.

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2 years ago

33.56 g of fructose (C6H,206) and 18.88 g of water are mixed to obtain a 40.00 ml solution a. What is this solution's density? b

Darina [25.2K]

Explanation:

Mass of fructose = 33.56 g

Mass of water = 18.88 g

Total mass of the solution = Mass of fructose + Mass of water = M

M = 33.56 g + 18.88 g =52.44 g

Volume of the solution = V = 40.00 mL

Density = $\frac{Mass}{Volume}$

a) Density of the solution:

$\frac{M}{V}=\frac{52.44 g}{40.00 mL}=1.311 g/mL$

b) Molar mass of fructose = 180.16 g/mol

Moles of fructose = $n_1=\frac{ 33.56 g}{180.16 g/mol}=0.1863 mol$

Molar mass of water = 18.02 g/mol

Moles of water= $n_2=\frac{ 18.88 g}{18.02 g/mol}=1.0477 mol$

Mole fraction of fructose in this solution: $\chi_1$

$\chi_1=\frac{n_1}{n_1+n_2}=\frac{0.1863 mol}{0.1863 mol+1.0477 mol}$

$\chi_1=0.1510$

Mole fraction of water = $\chi_2=1-\chi_1=0.8490$

c) Average molar mass of of the solution:

= $\chi_1\times 180.16 g/mol+\chi_2\times 18.02 g/mol$

$=0.1510\times 180.16 g/mol+0.8490\times 18.02 g/mol=42.50 g/mol$

d) Mass of 1 mole of solution = 42.50 g/mol

Density of the solution = 1.311 g/mL

d) Specific molar volume of the solution:

$\frac{\text{Average molar mass}}{\text{Density of the mass}}$

$=\frac{42.50 g/mol}{1.311 g/mL}=32.42 mL/mol$

5 0

3 years ago

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction

Tasya [4]

Answer:

0.456M

Explanation:

1. Balanced molecular equation

$2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O$

2. Mole ratio

$\dfrac{2molHNO_3}{1molBa(OH)_2}$

3. Moles of HNO₃

Number of moles = Molarity × Volume in liters
n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

4. Moles Ba(OH)₂

n = 0.700M × 0.0310 liter = 0.0217 mol

5. Limiting reactant

Actual ratio:

$\dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28$

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

6. Final molarity of Ba(OH)₂

Molarity = number of moles / volume in liters
Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M

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3 years ago

Can someone help pls

zysi [14]

What is the problem u need help with

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2 years ago

Explain the three types of cumbusion ??​

Explain the three types of cumbusion ??