Answer:
2KOH(aq) + NiSO₂(aq) → K₂SO₄(aq) + NiOH₂(s)
Explanation:
This reaction is an example of a <em>double-replacement reaction </em>where the cations of two compounds exchange with its anions. In the reaction:
KOH(aq) + NiSO₄(aq)
There are produced K₂SO₄ and NiOH₂ salts (The last one is insoluble, its state is (s) but K₂SO₄ is very soluble, its state is (aq). The unbalanced reaction is:
KOH(aq) + NiSO₄(aq) → K₂SO₄(aq) + NiOH₂(s)
To balance the potassiums:
<h3>
2KOH(aq) + NiSO₂(aq) → K₂SO₄(aq) + NiOH₂(s)</h3>
And now, the reaction is balanced
The first one is d and the second is b i think a nd. the last one is b too
Can you please translate it to english?
Answer:fH = - 3,255.7 kJ/mol
Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
When I change it, I think it's 5 mm