Answer:
The frequency of photon is 0.75×10¹⁵ s⁻¹.
Explanation:
Given data:
Energy of photon = 5×10⁻¹⁹ J
Frequency of photon = ?
Solution:
Formula;
E = hf
h = planck's constant = 6.63×10⁻³⁴ Js
5×10⁻¹⁹ J = 6.63×10⁻³⁴ Js ×f
f = 5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js
f = 0.75×10¹⁵ s⁻¹
The frequency of photon is 0.75×10¹⁵ s⁻¹.
An element is a pure substance made up of only one kind of atom. An element is composed of atoms in which it is composed of neutrons and protons. The nucleus is surrounded by electrons. The answer to this problem is A. element
Answer:
Identify one disadvantage to each of the following models of electron configuration:
Dot structures
Arrow and line diagrams
Written electron configurations
Explanation:
Identify one disadvantage to each of the following models of electron configuration:
Dot structures
Arrow and line diagrams
Written electron configurations
Answer:
Explanation:
formula of osmotic pressure is as follows
p= n RT
n is mole of solute per unit volume
If m be the grams of solute needed
m gram = m / 227.1 moles
m / 227.1 moles dissolved in .279 litres
n = m / (227.1 x .279 )
= m / 63.36
substituting the values in the osmotic pressure formula
5.14 = (m / 63.36) x .082 x 298
m / 63.36 = .21
m = 13.32 grams .
Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g
