A single replacement or displacement reaction occurs when one ingredient in a compound replaces another. Only another metal can replace a less active one; a metal can only displace another metal, and a non-metal can only displace another non-metal.
<h3>Displacement reaction:</h3>
The tendency of the element to lose electrons is what makes it metallic. Due to their bigger sizes and propensity to lose electrons as size, the number of shells, and nuclear force of attraction rise, the most active metals are found on the left side of the periodic table.
The reaction in which the more reactive element displaces the less reactive element is known as the displacement reaction.
When we combine an active metal with HCl, the reaction is as follows:
We are taking sodium, Na which is the most active metal. It will replace H in HCl because it is more reactive than H.
2Na + 2HCl → 2NaCl + H₂
Three ways to recognize if a single displacement reaction has taken place:
1. Color change
2. Formation of bubbles (a gas)
3. Formation of a precipitate
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Answer:
Moving them farther apart
Explanation:
One can try to change the distance between the two positive charges in such a way that it increases the distance and decreased the electric force.
Answer:
A) 9.60 g
B) 19.2 g
C) 28.8 g
Explanation:
A) Elemental oxygen mass is 16.00 g/mol, so in 0.600 mol of elementar oxygen there is 9.60 g
0.600 mol x 16.00 g/mol=9.60 g
B) Oxygen molecular mass is 32.00 g/mol, so in 0.600 mol of oxygen molecules there is 19.2 g
0.600 mol x 32.00 g/mol=19.2 g
C) Ozone molecular mass is 48.00 g/mol, so in 0.600 mol of oxygen molecules there is 28.8 g
0.600 mol x 48.00 g/mol=28.8 g
Answer: The enthalpy change for this reaction is, -803 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{CH_4}\times \Delta H_{CH_4})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BCH_4%7D%5Ctimes%20%5CDelta%20H_%7BCH_4%7D%29%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times -394)+(2\times -242]-[(2\times 0)+(1\times -75)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-394%29%2B%282%5Ctimes%20-242%5D-%5B%282%5Ctimes%200%29%2B%281%5Ctimes%20-75%29%5D)
Therefore, the enthalpy change for combustion of methane is, -803 kJ
Answer:it is necessary to use colourless
Explanation: A pure substance generally has a melting range (the difference between the temperature where the sample starts to melt and the temperature where melting is complete) of one or two degrees. Impurities tend to depress and broaden the melting range so the purified sample should have a higher and smaller melting range than the original, impure sample
