(NH(4))(2) Cr(2)O(7) = N(2) + H(2) O + C...

`(NH_(4))_(2) Cr_(2)O_(7) = N_(2) + H_(2) O + Cr_(2) O_(3)`

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To balance the redox reaction \( (NH_4)_2Cr_2O_7 \rightarrow N_2 + H_2O + Cr_2O_3 \), we will follow these steps: ### Step 1: Identify the oxidation states - In \( (NH_4)_2Cr_2O_7 \): - Nitrogen (N) in \( NH_4^+ \) has an oxidation state of -3. - Chromium (Cr) in \( Cr_2O_7^{2-} \) has an oxidation state of +6. - In the products: ...

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Intramolecular redox (NH_(4))_(2)Cr_(2)O_(7) rarr N_(2) + Cr_(2)O_(3) + 4H_(2)O Shows a gain and loss of how many electrons

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(NH_(4))_(2)Cr_(2)O_(7) overset(Delta)to N_(2)uarr+Cr_(2)O_(3)darr+H_(2)O uarr

What happens when (i) (NH_(4))_(2) Cr_(2) O_(7) is heated ? (ii) H_(3)PO_(3) is heated ? Write the equation

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RC MUKHERJEE-OXIDATION NUMBER AND BALANCING OF REDOX REACTIONS -PROBLEMS

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