Reaction equations contains symbols which show the physical state of the reactants and products.
<h3>What is a chemical equation:</h3>
A chemical equation is an expression that shows the interaction of reactants to yield products. Usually, symbols such as (s), (l), (g), and (aq) are used to show the state of the reactants and products.
The following are the respective meanings of these symbols;
- (s) - solid
- (l) - Liquid
- (g) - gas
- (aq) - dissolved in water
Learn more about reaction equations: brainly.com/question/1170557
Answer:
Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.
Explanation:
Answer:
1.64x10⁻¹⁸ J
Explanation:
By the Bohr model, the electrons surround the nucleus of the atom in shells or levels of energy. Each one has it's energy, and the electron doesn't fall to the nucleus because it can reach another level of energy, and then return to its level.
When the electrons go to another level, it absorbs energy, and then, when return, this energy is released, as a photon (generally as luminous energy). The value of the energy can be calculated by:
E = hc/λ
Where h is the Planck constant (6.626x10⁻³⁴ J.s), c is the light speed (3.00x10⁸ m/s), and λ is the wavelength of the photon.
The wavelength can be calculated by:
1/λ = R*(1/nf² - 1/ni²)
Where R is the Rydberg constant (1.097x10⁷ m⁻¹), nf is the final orbit, and ni the initial orbit. So:
1/λ = 1.097x10⁷ *(1/1² - 1/2²)
1/λ = 8.227x10⁶
λ = 1.215x10⁻⁷ m
So, the energy is:
E = (6.626x10⁻³⁴ * 3.00x10⁸)/(1.215x10⁻⁷)
E = 1.64x10⁻¹⁸ J
Answer:
NaNO3 (solubility = 89.0 g/100 g H2O)
Explanation:
The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.
Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)
Molar mass of NaNO3= 23+14+3(16)= 85gmol-1
Mass of solute=89.0g
Amount of solute= mass of NaNO3/molar mass of NaNO3
Amount of solute= 89.0g/85.0 gmol-1
= 1.0moles of NaNO3
Note that 100g of water=100cm^3 of water.
If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,
x moles of NaNO3 will dissolve in 1000cm^3 of water
x= 1.0 × 1000/ 100
x= 10.0 moles of NaNO3