Answer:
Percent ionic character of HI bond is 4.91%.
Explanation:
<h3>
Given Data:</h3>
Measured Dipole = 0.380D
bond distance = d = 161pm = 1.61*10^-8 cm
<h3>
Calculation:</h3>
% ionic character is determined by following equation:
% ionic= (dipole measured/dipole calculated)*100
Now,
(In above step 3*10^8 is multiplied to convert coulomb into esu)
As,
So,
Now we can % ionic character using above equation:
%ionic=(0.380D/7.728D)*100
% ionic character=4.91%
A) acids because they start with h
Answer:
44.63g
Explanation:
First, let us calculate the number of mole of KBr in 1.50M KBr solution.
This is illustrated below:
Data obtained from the question include:
Volume of solution = 250mL = 250/1000 = 0.25L
Molarity of solution = 1.50M
Mole of solute (KBr) =.?
Molarity is simply mole of solute per unit litre of solution
Molarity = mole /Volume
Mole = Molarity x Volume
Mole of solute (KBr) = 1.50 x 0.25
Mole of solute (KBr) = 0.375 mole
Now, we calculate the mass of KBr needed to make the solution as follow:
Molar Mass of KBr = 39 + 80 = 119g/mol
Mole of KBr = 0.375 mole
Mass of KBr =?
Mass = number of mole x molar Mass
Mass of KBr = 0.375 x 119
Mass of KBr = 44.63g
Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution
If element X has 59 protons then element X has 59 electrons.
Side note: as long as an element stays an atom the number of protons and electrons will always have the same value.
