AgNO3 reacts with CoCl2 based on the following equation:
<span>AgNO3 + CoCl2 ........> CoNO3 + AgCl<span>2
</span></span>
The complete ionic equation for this reaction is:
<span>Co2+(aq) + 2Cl- (aq) + 2Ag+(aq) + 2NO3-(aq) ...> 2AgCl(s) + Co2+(aq) + 2NO3-(aq)
</span>while the net ionic equation is:
<span>Cl-(aq) + Ag+(aq) → AgCl(s)
</span>
From this ionic equations, we can see that Co2+ and NO3- remained unchanged in the ionic equations. Therefore <span>Co2+ and NO3- are the spectator ions.</span>
Answer:
6633 grams of M are dissolved in 9L
Explanation:
If the compound M has a solubility in acetone of 0.737 g/mL, means that in 1 mLof solution, 0.737 grams of solute is dissolved.
Let's make a rule of three:
9 L = 9000 mL
In 1 mL ___ 0.737 g of M is dissolved
In 9000 mL ___ (9000 . 0.737) /1 = 6633 grams
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
