The formula is m = D x V
D = <span>13.69 g/cm^3.
</span>V = <span>15.0 cm^3
the mass of the liquid mercury is m= </span>13.69 g/cm^3 x 15.0 cm^3 = 195g
the molar mass of Hg is 200,
1 mole of Hg = 200g Hg, so #mole of Hg= 195 / 200 = 0.97 mol
but we know that
1 mole = 6.022 E23 atoms
0.97 mole=?
6.022 E23 atoms x 0.97 / 1 mole = 5.84 E23 atoms
Explanation:
According to Bohr's postulates, the electron in the present in the lower energy level can absorb energy and exits to higher energy level. Also, when this electron returns back to its orbit, it emits some energy.
Since the hydrogen consists of 1 electron and 1 proton. The lowest energy configuration of the hydrogen is when n =1 or, when the electron is present in the K-shell or the ground state.
The possible transition for the electron given in the question is :
n = 2, 3 and 4
The schematic diagram of the hydrogen atom consisting of these four quantum levels in which the electron can jump (Absorption) and comeback to from these energy levels (emission) .
Answer:
option (B) is correct
Explanation:
In case of nuclear reactors first the nuclear energy is emitted due to the nuclear fission of heavy elements.
This nuclear energy is emitted in the form of heat energy.
This heat energy is used to rotate the turbines, that means it is converted in the form of mechanical energy and then finally this mechanical energy is converted into electrical energy.
Answer:
Option D. 230 J
Explanation:
We'll begin by calculating the temperature change of the iron. This can be obtained as follow:
Initial temperature (T₁) = 50 °C
Final temperature (T₂) = 75 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 75 – 50
ΔT = 25 °C
Thus, the temperature change of the iron is 25 °C.
Finally, we shall determine the amount of heat energy used. This can be obtained as follow:
Mass (M) = 20 g
Change in temperature (ΔT) = 25 °C
Specific heat capacity (C) = 0.46 J/gºC
Heat (Q) =?
Q = MCΔT
Q = 20 × 0.46 × 25
Q = 230 J
Thus, the amount of heat used was 230 J
Answer:
Any of the six chemical elements that markup group1
of the periodic table.
Explanation: