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suter [353]
3 years ago
6

Research three different types of fire extinguisher and explain how these can be used to put out different fires, you must relat

e this to the fire triangle and the section that is removed
Chemistry
1 answer:
Vilka [71]3 years ago
7 0
The fire triangle is composed of 
1) heat
2) fuel
3) oxidizing agent (oxygen)

Types of fire extinguishers:
1) Water and Foam - <span> extinguish the fire by taking away the </span>heat<span> element of the fire triangle. Foam agents separate the </span>oxygen<span> element from the other elements.</span>
2) Carbon dioxide - <span>extinguish fire by taking away the </span>oxygen<span> element of the fire triangle and also be removing the </span>heat<span> with a very cold discharge.</span>
3) Dry Chemical - <span>extinguish the fire primarily by interrupting the </span>chemical reaction<span> of the fire triangle.</span>
4) Wet Chemical - <span>extinguishes the fire by removing the heat of the fire triangle and prevents re-ignition by creating a barrier between the </span>oxygen<span> and </span>fuel<span> elements</span>
5) Clean Agent - <span>extinguish the fire by interrupting the </span>chemical reaction<span> of the fire triangle</span>
6) Dry Powder - extinguish the fire by separating thefuel<span> from the </span>oxygen<span> element or by removing the </span>heat<span> element of the fire triangle</span>
7) Water Mist - <span>extinguish the fire by taking away the </span>heatelement of the fire triangle
8) Cartridge Operated Dry Chemical - extinguish the fire primarily by interrupting the chemical reaction<span> of the fire triangle</span>

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Top fuel dragsters and funny cars burn nitro-methane as fuel according to the following balanced combustion equation: 2CH3NO2(l)
mart [117]

Answer : The standard enthalpy of formation for nitro-methane is, -467.4 kJ

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

The equation used to calculate enthalpy change is of a reaction is:  

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

The equilibrium reaction follows:

2CH_3NO_2(l)+\frac{3}{2}O_2(g)\rightleftharpoons 2CO_2(s)+3H_2O(g)+N_2(g)

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

\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})+(n_{(N_2)}\times \Delta H^o_f_{(N_2)})]-[(n_{(CH_3NO_2)}\times \Delta H^o_f_{(CH_3NO_2)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]

We are given:

\Delta H^o_{rxn}=-709.2kJ/mol

\Delta H^o_f_{(CH_3NO_2(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-286kJ/mol\\\Delta H^o_f_{(N_2(g))}=0kJ/mol

Putting values in above equation, we get:

-709.2=[(2\times -393)+(3\times -286)+(1\times 0)]-[(2\times \Delta H^o_f_{(CH_3NO_2)})+(\frac{3}{2}\times 0)]

\Delta H^o_f_{(CH_3NO_2)}=-467.4kJ

Thus, the standard enthalpy of formation for nitro-methane is, -467.4 kJ

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