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dolphi86 [110]
1 year ago
12

When a piece of metal A is placed in a solution containing ions of metal B, metal B plates out on the piece of A.(a) Which metal

is being oxidized?
Chemistry
1 answer:
zmey [24]1 year ago
5 0

While metal B is being reduced from an ion to a solid metal, metal A is being oxidized.

Metal A is being displaced because, through oxidation, it transforms from a solid metal to ions in a solution.

What is Oxidation ?

Redox reactions include a change in the oxidation state of the substrate. Loss of electrons or a rise in an element's oxidation state are both considered to be oxidation.

Gaining electrons or lowering the oxidation state of an element or its constituent atoms are both examples of reduction.

In general, the term "reduction" refers to a reaction in which a chemical receives additional electrons; the compound that gets electrons is referred to as being reduced. We can think of the transformation of a ketone or an aldehyde into an alcohol as a two-electron reduction because hydride can be thought of as a proton plus two electrons.

So finally we can say that Metal A is being oxidized.

To know more about Oxidation please click here : brainly.com/question/25886015

#SPJ4

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What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should
horsena [70]

Answer:

i

J_{m} = 20

ii

J_{m} = 22.5

Explanation:

From the question we are told that

  The first temperatures is T_1 =  25^oC =  25 +273 =298 \ K

   The second temperature is  T_2 =  100^oC =  100 +273 = 373 \ K

Generally the equation for  the most highly populated rotational energy level is mathematically represented as

     J_{m} = [ \frac{RT}{2B}]  ^{\frac{1}{2} } - \frac{1}{2}

Here R is the gas constant with value R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}

Also  

      B is given as B=\ 0.244 \ cm^{-1}

   Generally the energy require per mole to move 1 cm is  12 J /mole

So   0.244 \ cm^{-1}  will require x J/mole

           x =  0.244 *  12

=>          x =  2.928 \ J/mol

So at the first temperature

     J_{m} = [ \frac{8.314 * 298  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5

=>  J_{m} = 20

So at the second temperature

           J_{m} = [ \frac{8.314 * 373  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5

=>  J_{m} = 22.5

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