Answer:
True
Explanation:
The physical changes are reversible in most cases and these changes are not the chemical changes which means that it is only the change in its state not in their nature. Just take the example of water, on cooling it becomes solid and change in color can be seen which is white in solid form and colorless in liquid form. This is also reversible and is a physical change. This means that physical changes can be identified at macroscopic level. Hence the answer is true.
valence electrons are the number of electrons in the outer shell. there can only be 8 electrons in the outer shell. The number of valence electrons can be used to determine how many bonds are needed.
For example: H2O
O (oxygen) has 6 valence electrons
H (hydrogen) has 1 valence electron
O needs 2 more electrons to be stable
H needs 1 more electron to be stable
O forms one bond with two H atoms to form H2O.
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
H2SO4 + 2RbOH -> Rb2SO4 + 2H2O
If you want an explanation, keep reading.
In the first portion, there are two hydrogen ions and four sulfate ions.
The second portion has one rubidium ions and one hydroxide ion.
On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.
Back to the right side, there is there is water (H2O).
On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).
So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).
I hope this was easy to understand.
