= 9.1 × 10^6
(scientific notation)
= 9.1e6
(scientific e notation)
= 9.1 × 10^6
(engineering notation)
(million; prefix mega- (M))
= 9100000
<span>(real number)</span>
Answer:
Explanation:
From the question we are told that:
Slope 
initial Concentration 
Time 
Generally the equation for Raw law is mathematically given by
Answer:
N2 + H2 ----------》NH3
On balancing it
N2. + 3.H2------->2.NH3
( 1 mol) (3 mol) (2 mol)
1 mol of nitrogen reacts with 3 mol of hydrogen to give 2 mol of ammonia.
Likewise,
20 litres of nitrogen reacts with 60 litres of hydrogen to give 40 litres of Ammonia.
Hence, the answer is 40 Litres.
Answer:
Option C = electron
Explanation:
Electrons are responsible for the production of colored light.
Electron:
The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.
Symbol= e-
Mass= 9.10938356×10⁻³¹ Kg
It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.
How electrons produce the colored light:
Excitation:
When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.
De-excitation:
When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.
Other process may involve,
Fluorescence:
In fluorescence the energy is absorbed by the electron having shorter wavelength and high energy usually of U.V region. The process of absorbing the light occur in a very short period of time i.e. 10 ∧-15 sec. During the fluorescence the spin of electron not changed.
The electron is then de-excited by emitting the light in visible and IR region. This process of de-excitation occur in a time period of 10∧-9 sec.
Phosphorescence:
In phosphorescence the electron also goes to the excitation to the higher level by absorbing the U.V radiations. In case of Phosphorescence the transition back to the lower energy level occur very slowly and the spin pf electron also change.
Answer:
992.302 K
Explanation:
V(rms) = 750 m/s
V(rms) = √(3RT / M)
V = velocity of the gas
R = ideal gas constant = 8.314 J/mol.K
T = temperature of the gas
M = molar mass of the gas
Molar mass of CO₂ = [12 + (16*2)] = 12+32 = 44g/mol
Molar mass = 0.044kg/mol
From
½ M*V² = 3 / 2 RT
MV² = 3RT
K = constant
V² = 3RT / M
V = √(3RT / M)
So, from V = √(3RT / M)
V² = 3RT / M
V² * M = 3RT
T = (V² * M) / 3R
T = (750² * 0.044) / 3 * 8.314
T = 24750000 / 24.942
T = 992.302K
The temperature of the gas is 992.302K
Note : molar mass of the gas was converted from g/mol to kg/mol so the value can change depending on whichever one you use.
