Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent Bond
Between 0.4 and 1.7 then it is Polar Covalent Bond
Greater than 1.7 then it is Ionic
For Br₂;
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar Covalent Bond)
For MgS;
E.N of Sulfur = 2.58
E.N of Magnesium = 1.31
________
E.N Difference 1.27 (Ionic Bond)
For SO₂;
E.N of Oxygen = 3.44
E.N of Sulfur = 2.58
________
E.N Difference 0.86 (Polar Covalent Bond)
For KF;
E.N of Fluorine = 3.98
E.N of Potassium = 0.82
________
E.N Difference 3.16 (Ionic Bond)
Result: The Bonds in Br₂ and SO₂ are Covalent in Nature.
Answer:
carbon, nitrogen,oxygen,flourine
Answer:
a
Explanation:
it is A because h20 and c02 are molecules while what make them up are atoms
Answer:
Neon
Explanation:
Step 1: Given and required data
- Density of the gas (ρ): 1.57 g/L
- Ideal gas constant (R): 0.08206 atm.L/mol.K
Step 2: Convert T to Kelvin
We will use the following expression.
K = °C + 273.15 = 40.0 + 273.15 = 313.2 K
Step 3: Calculate the molar mass of the gas (M)
For an ideal gas, we will use the following expression.
ρ = P × M/R × T
M = ρ × R × T/P
M = 1.57 g/L × 0.08206 atm.L/mol.K × 313.2 K/2.00 atm
M = 20.17 g/mol
The gas with a molar mass of 20.17 g/mol is Neon.
Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:

a) moles of butane

b) moles of oxygen


According to stoichiometry :
2 moles of butane require 13 moles of 
Thus 0.09 moles of butane will require =
of 
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.