Answer:
- The answer is the concentration of an NaOH = 1.6 M
Explanation:
The most common way to solve this kind of problem is to use the formula
In your problem,
For NaOH
C₁ =?? v₁= 78.0 mL = 0.078 L
For H₂SO₄
C₁ =1.25 M v₁= 50.0 mL = 0.05 L
but you must note that for the reaction of NaOH with H₂SO₄
2 mol of NaOH raect with 1 mol H₂SO₄
So, by applying in above formula
- (C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)
- C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M
<u>So, the answer is the concentration of an NaOH = 1.6 M</u>
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
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For C and N,
E.N of Nitrogen = 3.04
E.N of Carbon = 2.55
________
E.N Difference 0.49 (Weakly Polar Covalent)
For N and S,
E.N of Nitrogen = 3.04
E.N of Sulfur = 2.58
________
E.N Difference 0.46 (Weakly Polar Covalent)
For K and F,
E.N of Fluorine = 3.98
E.N of Potassium = 0.28
________
E.N Difference 3.70 (Ionic)
For N and N,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference 0 (Non Polar Covalent)
Answer:
Here's what I find.
Explanation:
Many scientists contributed to our model of the atom.
Among those who received the Nobel Prize for their work are:
1906 — J.J. Thomson — discovery of the electron
1908 — Ernest Rutherford — nuclear model of the atom
1922 — Niels Bohr — planetary model of the atom
1922 — Albert Einstein — quantum mechanical model of the atom
1935 — James Chadwick — discovery of the neutron
