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In short, Your Answer would be Option C
Hope this helps!</span>
The reactions are in order which includes combustion reaction, Hydration reaction, oxidation reaction, and displacement reaction.
a) A combustion reaction is a chemical reaction between a fuel and an oxidant where heat is released. The combustion reaction example is given below. It is a balanced chemical reaction.
2C₃H₆(g) + 9O₂(g) --------> 6CO₂(g) + 6H₂O(g)
b. A hydration reaction is a chemical reaction in which a molecule of water is added to another molecule. Here Aluminum oxide is added to water to form aluminum hydroxide.
4Al₂O3(s) + 6H₂O(l)------> 2Al(OH)3(s)
c. When a metal reacts with oxygen, the metal forms an oxide. Oxide is a compound of metal and oxygen. Here lithium metal reacts with oxygen to form lithium oxide.
2Li(s) + O₂(g)-----> Li₂O(s)
d. A displacement reaction is one in which a more reactive element displaces a less reactive element from a compound. Here Zinc is more reactive than silver, so silver was displaced to form Zinc Nitrate.
Zn(s) + 2AgNO₃(aq) -----> 2Ag(s) + Zn(NO₃)₂(aq)
To know more about reactions, click below:
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Answer:
The particle theory is used to explain the properties of solids, liquids and gases. The strength of bonds (attractive forces) between particles is different in all three states.
Explanation:
Answer:
sorry I have no idea what the answer is
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
