n / V = P / RT = (4.77 atm) / ((0.08205746 L atm/K mol) x (118 K)) = 0.4926 mol/L
(2.016 g/mol) x (0.4926 mol/L) = 0.993 g/L
20 Ca :
1s², 2s², 2p⁶, 3s², 3p⁶, 4s²<span>
</span>
Answer D
hope this helps!
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
Answer:
Oxidizing agent - CrO4^2-
Reducing agent- N2O
Explanation:
Let us look at the equation closely;
CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]
The reduction half equation is;
CrO4^2- (aq) + 3e -------->Cr^3+ (aq)
Oxidation half equation is;
3N2O(g) ------>3 NO(g) +3 e
Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.
The question is basically asking what is happening to the energy (that is in the form of heat) when it is being absorbed by an object. The energy being absorbed from the heat source is being turned into kinetic energy. This can be explained by temperature change. As you add more heat to an object, the temperature rises. Since temperature is the average kinetic energy of all of the molecules in an object, we can say that as temperature rises so does the kinetic energy of the molecules in the object. Due to the fact that heat is causing the temperature to increase, we can say that the energy from the heat is being turned into kinetic energy.
I hope this helps. Let me know in the comments if anything is unclear.
