The reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
<h3>What is metal displacement reaction? </h3>
Displacement reactions is a reaction which includes a metal and the compound of a other metal. A more reactive metal will push or displace out a less reactive metal from its compound in this displacement reaction. The metal which is less reactive left uncombined after the reaction.
As we know that, electrons are the basis of the chemical reactions. If chemical compound or element A is more easily oxidized than B, then according to the terms of the activity series, the elements which are more easily oxidized can react with more chemicals, since they are able to act as a reducing agents for more chemicals.
Since, Metal ions are positively charged ions as they lose electrons. Some metals give up their electrons more readily than others and become more reactive.
Thus, we concluded that the reduction of a less active metal by a more active one is called metal displacement reactions. For example:
Fe + CuSO4 → FeSO4 + Cu
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we are given the the two reactants: AgNO3 and Na2CO3 and is asked to write a balanced equation and a net ionic equation for the reaction of the two. This is a double-replacement reaction:
2AgNO3 (aq)+ Na2CO3 (aq)= Ag2CO3 + 2NaNo3 (aq)
2 Ag + + 2 N03- + 2Na+ + CO32- = Ag2CO3 + 2 Na+ 2NO3-
cancelling the spectator ions, 2Ag + + CO32- = Ag2CO3
C) It contains the same number of electrons and protons.
Answer:
0.42 M
Explanation:
The reaction that takes place is:
- Cu(CH₃COO)₂ + Na₂CrO₄ → Cu(CrO₄) + 2Na(CH₃COO)
First we <u>calculate the moles of Na₂CrO₄</u>, using the <em>given volume and concentration</em>:
(200 mL = 0.200L)
- 0.70 M * 0.200 L = 0.14 moles Na₂CrO₄
Now we <u>calculate the moles of Cu(CH₃COO)₂</u>, using its <em>molar mass</em>:
- 40.8 g ÷ 181.63 g/mol = 0.224 mol Cu(CH₃COO)₂
Because the molar ratio of Cu(CH₃COO)₂ and Na₂CrO₄ is 1:1, we can directly <u>substract the reacting moles of Na₂CrO₄ from the added moles of Cu(CH₃COO)₂</u>:
- 0.224 mol - 0.14 mol = 0.085 mol
Finally we <u>calculate the resulting molarity</u> of Cu⁺², from the <em>excess </em>cations remaining:
- 0.085 mol / 0.200 L = 0.42 M