Answer:
a) [Tris0] : [Tris] = 1 : 100
b) Range = 7.1 to 9.1
Explanation:
a) Calculation of ratio of the basic and the acidic forms of tris
pH of a buffer is calculate using Henderson-Hasselbalch equation
Conjugate acid of Tris dissociated as
For tris,
Salt or Basic form = tris0
Acid or Acidic form = Tris
pKa = 8.1
pH = 6.1
[Tris0] : [Tris] = 1 : 100
b) Range of Tris
Range of any buffer is:
From (pKa -1) to (pKa+1)
So, range of Tris is:
From (8.1 - 1) to (8.1 +1)
or from 7.1 to 9.1
Answer:
0.241 × 10³⁰ molecules
Explanation:
Given data:
Mass of Cr(HCO₃)₃ = 9.273 × 10⁷ g
Number of molecules = ?
Solution:
Number of moles = 9.273 × 10⁷ g/ 235 g/mol
Number of moles = 0.04× 10⁷ mol
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
For 0.04× 10⁷ moles of Cr(HCO₃)₃:
0.04× 10⁷ moles × 6.022 × 10²³ molecules / 1 mol
0.241 × 10³⁰ molecules
Answer:
a
Explanation:
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Chemical Formula of Sodium Sulfate is Na₂SO₄.
As we know, 1 mole of any substance contains 6.022 × 10²³ particles. So, 1 mole of Na₂SO₄ also contains 6.022 × 10²³ formula units of Na₂SO₄.
Also,
1 mole of Na₂SO₄ contains = 6.022 × 10²³ Atoms of S
So,
4.5 moles of Na₂SO₄ will contain = X atoms of S
Solving for X,
X = (4.5 moles × 6.022 × 10²³ Atoms) ÷ 1 mole
X = 2.70 × 10²⁴ Atoms of S
Result:
4.5 Moles of Na₂SO₄ contains 2.70 × 10²⁴ Atoms of Sulfur.
The mass of titanium is = 47,867 g/1mol
Applying the rule of avrogado
1mol _______ 6,023 × 10^(23) at
0,075mol ___ x
X . 1mol = 0,075mol . 6,023 . 10^(23)at
X = 0,075 . 6,023 . 10^(23) at
X = 4,51 . 10^(22) atoms
Hope this helps
