There are several scales to determine the weight of material. The chose of the weight scale depends upon the size of the material. Like if we have to measure the molecular or atomic weight of a molecule or atom we use gm. Like that for the bigger molecules like proteins the weight of the molecule is expressed by different scale due to its big size. Like that for big material like iron rods etc we generally use kilograms or ton. There are two types of measuring units one is C.G.S. and another is S.I. The S.I. units are more acceptable worldwide than that the C.G.S. units as the digit comes in SI units are easily handable than that of CGS.
Thus metric is used to determine a large weight of compounds which is a SI unit and handable.
Answer:
(2) The lowest energy orbits are those closest to the nucleus.
Explanation:
In the Bohr theory the electrons describe circular orbits around the nucleus of the atom without radiating energy, therefore to maintain the circular orbit, the force that the electron experiences, that is, the coulombian force due to the presence of the nucleus, must be equal to the centripetal force.
The electron only emits or absorbs energy in the jumps from one allowed orbit to another, with only one jump occurring at a time, from layer K (n = 1) to layer L (n = 2), without going through intermediate orbits. In said change it emits or absorbs a photon whose energy is the difference in energy between both levels.
In Bohr's model, it is stipulated that the energy of the electron is greater the greater the radius r, so the lowest energy orbits are those closest to the nucleus.
Answer:
6.022×1023 individual items of that something. You have 3 moles, so there are 3×6.022×1023 oxygen molecules
Explanation:
6.022×1023 individual items of that something. You have 3 moles, so there are 3×6.022×1023 oxygen molecules
Answer:
Because you have to give the simplest ratio of equation. Equation I can be simplified by dividing all the coefficients by 2 to make it into equation II.
*Every equation has to be in the simplest form of coefficients.
Ratio of Equation I :
2 : 4 : 2 : 4
Ratio of Equation II :
1 : 2 : 1 : 2
Answer:
is the change in enthalpy associated with the combustion of 530 g of methane.
Explanation:
Mass of methane burnt = 530 g
Moles of methane burnt = 
Energy released on combustion of 1 mole of methane = -890.8 kJ/mol
Energy released on combustion of 33.125 moles of methane :
is the change in enthalpy associated with the combustion of 530 g of methane.
