Chemistry Forum

In the exercise K 2.2* 10^-5 mol^¹* s^-1 and time 5 h is given.

Calculate the moles consumed = time × K

We have 1 mol NOBr - consumed moles = moles left

1. 2 Na3PO4 + 3 Ba(NO3)2 => 2 Ba3(PO4)2 + 6 NaNO3

2. 2 H3AsO4 + 3 Pb(NO3)2 => 6 HNO3 + 2 Pb3(AsO4)2
Or
H3AsO4 + Pb(NO3)2 => 2 HNO3 + PbHAsO4

It said A2 is reduced 2 times, not consummed. So it was taken out or removed. The same has to be done with B2.
The reaction is of course 1:1.
But if 0.02 mol A2 has reacted, then also 0,02 mol B2 has reacted and the new value is 0,03 mol.

I think its not a consumption.
The concentration is reduced from 0,04 to 0,02
The reaction took place 0,04 with 0,05 means to keep the ratio it is 0,02 to 0,025. B is also reduced 2 times from 0,05 to 0,025.

Yes I would think so

K = cPCl5/(cPCl3*cCl2)

4 = x/((3-x)*(2-x)) solve for x.

The other reaction is
If 1,5 mol/ l Cl2 is left then 0,5 mol/l has reacted. What means the same amount of PCl3 has reacted and also the same amount PCl5 was obtained.

K= 0,5/((3-0,5)*(2-0,5))

The 3.2 is removed from 7.2 to get final 4.

2D + G => 2C + B In equilibrium we have 4 mol/l D , 2 mol/l G and 3,2 mol/l C In beginning there is no C and no B. So it means 3,2 mol C are produced, what means 2 D gives 2 C. 3,2 mol C has to be add to D what gives 7,2 mol/l . 3,2 mol C means also the half of G = 1,6 mol/l has to be used. So in be...

It is alcaline, because the OH group in NaOH is bonded to a metal. If OH group is bonded to a nonmetal like C, N, S, P, Cl and others it is working neutral to acidic, depending on the electronegativity.
More alcaline is C2H5O- in C2H5ONa.

It is because of the OH- group. NaOH, KOH, Ca(OH)2, Ba(OH)2 and others have same behaviour.
The OH- group can react with H+ ( H3O+) to form H2O. It is a proton acceptor (Brönstedt) or a electron pair donator ( Lewis).

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Separate the Masses of CuSO4 and Water. With the specific gravity of the 80 ml solution you get the mass of it with percentage the mass of CuSO4. Calculate the mass of CuSO4 from the 30 g CuSO4×5 H2O and also the mass of the 5 H2O. So you will get two sum, one of the both CuSO4 amounts and the water...

Disaggree The solution is NiSO4. In this Chlorine Cl2 will be inserted. We have Ni 2+, SO4 2- and Cl2. There is no reaction Ni => Ni 2+ + 2e- Cl2 + 2e => 2Cl Ni 2+ + 2e => Ni This would be both reduction, no oxidation You cannot add both, because you need an Oxidation Ni => Ni 2+ + 2e- Cl2 + Ni => 2...

Its not clear do you have Ni metal available or not. If yes Ni is immersed into the NiSO4 solution. Chlorine is an oxidiser and will react with the Ni to form NiCl2. If only Ni 2+ is available then it could react from Ni 2+ to Ni 3+.