can someone help with this question?

[helloimpat]

Could someone help me with b? the suggest answer is 0.16.

Thanks!

[Sine]

yeah answer is correct. Since the equation is reversed we take the reciprocal of the initial K value so K(new)=1/K=1/6.3 = 0.16

[helloimpat]

Thank you

[helloimpat]

Can someone help me with another question?

[keltingmeith]

Can someone help me with another question?

Which question? What are you struggling with?

[helloimpat]

For question 12, i found the coloumbs (4420) from 5 times 884 and then found the mole of electrons = 0.045 mol (4420/96500). I'm not sure what to do next.

[AliTan]

Hi Pat,

With question 12, you have calculated that the number of coulombs that must have been passed through is 4420 and that the number of mole of electrons is 0.0458 mol (4420/96500). The question states that the mass of the cathode has increased from 2.428g to 3.267g meaning that 0.839g (3.267 - 2.428) of Manganese has been plated onto the cathode as Manganese is being reduced. From the data booklet, we know that the molar mass of Manganese is 54.9g/mol and thus the mol of Mn plated is 0.0153 mol (0.839/54.9). The ratio of the number of mol of Mn plated to the number of mol of electrons is 1:3 (0.0153:0.0458) so therefore the equation of the reduction reaction is Mn3+ + 3e- -> Mn. Therefore, the charge on the Manganese ion in the Nitrate Salt is 3+.

Hope this solved your question and good luck for year 12 VCE!