What is the oxidation number of chromium in k2cr2o7

The oxidation state of chromium in K₂Cr₂O₇ is 6⁺ (i.e. + 6).

Explanation:

You can calculate the oxidation number of most elements following some simple rules.

This is how you do it for chromium in K₂Cr₂O₇.

1) Rule: in a neutral compound the net oxidation number is zero (0).

Hence, sum of the oxidation numbers of K, Cr and O in K₂Cr₂O₇ is 0.

2) Rule: The most common oxidation number of oxygen in compounds, except in peroxides, is  2 ⁻ (negative 2).

3) Rule: the most common oxidation state of alkali metals is 1⁺ (positive 1)

4) Rule: multiply each oxidation state by the corresponding number of atoms in the compound (the subscripts)

2(1⁺) + 2(x) + 7(2⁻) = 0

          ↑         ↑       ↑

          K        Cr      O

2 + 2x - 14 = 02x - 12 = 02x = 12x = 6

Hence, the oxidation number of chromium in this compound is 6⁺.

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

Hello! data: p (pressure) = 1.25 atmv (volume) = ? (in l)t (temperature) = 298 kn (number of mols) = 0.900 molesr (gas constant) = 0.082 (atm*l/mol*k)apply the data to the clapeyron equation (ideal gas equation), see: i hope this . =)