Answer: The bond formed between the elements will be ionic bond.
Explanation: We are given two elements having electronic configurations:
Element 1: 
Element 2: 
Element 1 can easily loose 1 electron to attain stable electronic configuration and Element 2 can accept 1 electron to attain stable electronic configuration.
For these elements, there will be a complete transfer of electron from Element 1 to Element 2. Hence, this will form a ionic bond.
From the configuration, Element 1 is Lithium and Element 2 is Fluoride. So, the compound is LiF.
Answer:
31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .
Explanation:
Given data:
Number of moles of CaCl₂ = 2.6 mol
Number of Cl₂ ions = ?
Solution:
CaCl₂ → Ca²⁺ + 2Cl⁻
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
In one mole of CaCl₂ there are two moles of chloride ions present.
In 2.6 mol:
2.6×2 = 5.2 moles
1 mole Cl⁻ = 6.022 × 10²³ number of Cl⁻ ions
5.2 mol × 6.022 × 10²³ number of Cl⁻ / 1mol
31.31× 10²³ number of Cl⁻
Answer:
thomson developed the chocolate chip method which was the identification of the electrons in the core of an atom. Rutherford discovered that the core was only positive and that the electrons were floating outside of the core.
Explanation:
4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.
Solution:
We will start with the Molarity
Also we know 1000 ml = 1 L
Therefore 37.5 ml by 1000ml we obtained 0.0375L
Equation for solving mole of solute
Now, multiply 0.750M by 0.0375
Substitute the known values in the above equation we get
Also we know that Molar mass of KI is 166 g/mol
So divide the molar mass value to get the no of grams.
So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.
Answer: how much mass is in a certain amount of space
Explanation: density is g/mL; therefore it’s the mass (g) in a certain amount of space (mL)
