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:
Explanation:
When you divide exponentials, you subtract the powers. For the numbers infront, just use a basic calculator for.
7.95/6.02 = 1.32
10^22/10^23 = 10^-1
1.32 x 10^-1 is your answer
Answer:
Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.
Explanation:
The correct answer is option 2. A 0.8 M aqueous solution of NaCl has a higher boiling point and a lower freezing point than a 0.1 M aqueous solution of NaCl. This is explained by the colligative properties of solutions. For the two properties mentioned, the equation for the calculation of the depression and the elevation is expressed as: ΔT = -Km and <span>ΔT = Km, respectively. As we can see, concentration and the change in the property has a direct relationship.</span>
The number of mole of HCl needed for the solution is 1.035×10¯³ mole
<h3>How to determine the pKa</h3>
We'll begin by calculating the pKa of the solution. This can be obtained as follow:
- Equilibrium constant (Ka) = 2.3×10¯⁵
- pKa =?
pKa = –Log Ka
pKa = –Log 2.3×10¯⁵
pKa = 4.64
<h3>How to determine the molarity of HCl </h3>
- pKa = 4.64
- pH = 6.5
- Molarity of salt [NaZ] = 0.5 M
- Molarity of HCl [HCl] =?
pH = pKa + Log[salt]/[acid]
6.5 = 4.64 + Log[0.5]/[HCl]
Collect like terms
6.5 – 4.64 = Log[0.5]/[HCl]
1.86 = Log[0.5]/[HCl]
Take the anti-log
0.5 / [HCl] = anti-log 1.86
0.5 / [HCl] = 72.44
Cross multiply
0.5 = [HCl] × 72.44
Divide both side by 72.44
[HCl] = 0.5 / 72.4
[HCl] = 0.0069 M
<h3>How to determine the mole of HCl </h3>
- Molarity of HCl = 0.0069 M
- Volume = 150 mL = 150 / 1000 = 0.15 L
Mole = Molarity x Volume
Mole of HCl = 0.0069 × 0.15
Mole of HCl = 1.035×10¯³ mole
<h3>Complete question</h3>
How many moles of HCl need to be added to 150.0 mL of 0.50 M NaZ to have a solution with a pH of 6.50? (Ka of HZ is 2.3 x 10 -5 .) Assume negligible volume of the HCl
Learn more about pH of buffer:
brainly.com/question/21881762
