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:
<em>What is atmospheric pressure? -------> Atmospheric pressure is a force in an area pushed against a surface by the weight of the atmosphere of Earth, a layer of air.</em>
<em>Why does the atmosphere exert pressure? -------> Because gas particles in the air—like particles of all fluids—are constantly moving and bumping into things, so they exert pressure. </em>
<em>What is the value of atmospheric pressure at sea level, in newtons per square centimeter? -------> Atmospheric pressure at sea level is about 10 N/cm2 or 100 kPa or about 10 m of water or about 760 mm of mercury, but varies with the weather, and of course altitude.</em>
<em>I hope this helps and have a great day!</em>
Explanation:
Answer: 351
Explanation: From °C to kelvin we do
K=273+°C
From Kelvin to °C we do
°C=K-273 or -273+ K
Answer:
A third type of circuit involves the dual use of series and parallel connections in a circuit; such circuits are referred to as compound circuits or combination
Explanation:
Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
Regards.
