An electrolyte is a substance that releases ions when it is dissolved in water. Electrolytes are divided into salts, ions and bases. In this case, the electrolyte is H₃PO₄ as it releases H+ ions and PO₄⁻³ ions when it is dissolved in water.
Answer:
See explanation
Explanation:
According to the law of conservation of mass; the total mass of reactants on the left hand side of the reaction equation is equal to the total mass of products on the right hand side of the reaction equation.
Hence, the total mass of each atom on either side of the reaction equation should be exactly the same.
Since there are two atoms of oxygen on the reactants side, the total mass of oxygen = 16 amu * 2 = 32 amu
Since there are two oxygen atoms on the products side, total mass of oxygen = 16 amu * 2 = 32 amu
What mass of the following chemicals is needed to make the solutions indicated?
Answer:
271.6g
Explanation:
The mass of the chemicals need to make the needed solution can be derived by obtaining the number of moles first.
Given parameters:
Volume of solution = 1L
Molarity of HgCl₂ = 1M
number of moles of HgCl₂ = molarity of solution x volume
= 1 x 1
= 1 mole
From;
Mass of a substance = number of moles x molar mass;
we can find mass;
Molar mass of HgCl₂ = 200.6 + 2(35.5) = 271.6g/mol
Mass of the substance = 1 x 271.6 = 271.6g
Answer:
2.51 Angstroms
Explanation:
For a particle in a one dimensional box, the energy level, En, is given by the expression:
En = n²π² ħ² / 2ma²
where n is the energy level, ħ² is Planck constant divided into 2π, m is the mass of the electron ( 9.1 x 10⁻³¹ Kg ), and a is the length of the one dimensional box.
We can calculate the change in energy, ΔE, from n = 2 to n= 3 since we know the wavelength of the transition ( ΔE = h c/λ ) and then substitute this value for the expresion of the ΔE for a particle in a box and solve for the length a.
λ = 207 nm x 1 x 10⁻⁹ m/nm = 2.07 x 10⁻⁷ m ( SI units )
ΔE = 6.626 x 10⁻³⁴ J·s x 3 x 10⁸ m/s / 2.07 x 10⁻⁷ m
ΔE = 9.60 x 10⁻¹⁹ J
ΔE(2⇒3) = ( 3 - 2 ) x π² x ( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )
9.60 x 10⁻¹⁹ J = π² x( 6.626 x 10⁻³⁴ J·s / 2π )² / ( 2 x 9.1 x 10⁻³¹ Kg x a² )
⇒ a = 2.51 x 10⁻¹⁰ m
Converting to Angstroms:
a = 2.51 x 10⁻¹⁰ m x 1 x 10¹⁰ Angstrom / m = 2.51 Angstroms
Explanation:
- Simple Distillation: its a separation method that can be used when the two or more liquids in the mix have at least 50 degrees of difference between their boiling points.
-Azeotropic distillation: is a technique to break an azeotrope (constant boiling point mixtures), that can't be separated by simple distillation, by adding another component to generate a new azeotrope (between one initial component and the new one added) with lower boiling point.
-Extractive distillation: is a process to separate mixtures with close boiling points by adding a miscible, high boiling or none volatile solvent to increase the relative volatility of the liquids in the mix, this increases the separation factor. It differences from the azeotropic method because it doesn't form an azeotrope.
-Liquid-liquid extraction: is a method to separate compounds based on their relative solubilities in two different immiscible liquids.
After describing all the methods we can conclude that all of them are methods to separate substances based on their physical properties, this is their similarity. The difference between this method is the property it uses to separate (solubility in the case of extraction and boiling point in the case of destinations), the cases in which they bare used (when the liquids difference in boiling points is bigger [simple] or close [attractive and azeotropic]) and the formation of azeotropes (present in azeotropic and absent in extractive).
I hope you find this information useful and interesting! Good luck!
