The molarity of the acid sample H₂SO₄ is 0.052M .
<h3>What is Molarity ?</h3>
Molarity (M) is the amount of a substance in a certain volume of solution.
Molarity is defined as the moles of a solute per liters of a solution.
Molarity is also known as the molar concentration of a solution
Now to determine the molarity of the acid sample
V( H₂SO₄) = 24.0 mL in liters = 24.0 / 1000 = 0.024 L
M(H₂SO₄) = ?
V(NaOH) = 20.0 mL = 20.0 / 1000 = 0.02 L
M(NaOH) = 0.125 M
Number of moles NaOH :
n = M x V
n = 0.125 x 0.02
n = 0.0025 moles of NaOH
H₂SO₄(aq) + 2 NaOH(aq) = Na₂SO₄(aq) + 2 H₂O(l)
1 mole H₂SO₄ ---------- 2 mole NaOH
? mole H₂SO₄ ---------- 0.0025 moles NaOH
moles = 0.0025 * 1 / 2
= 0.00125 moles of H₂SO₄
M(H₂SO₄) = n / V
M = 0.00125 / 0.024
= 0.052 M
Therefore the molarity of the acid sample H₂SO₄ is 0.052M .
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Answer:
mass ratio of A/B is 2:1
Explanation:
Since the mass of box A = 10g
mass of box B = 5g
Mass of box C = mass of box A + mass of box
A ratio compares two quantities. To find the ratio of the two boxes:
Ratio of A to B = 
Ratio of A to B = 
= 2
The mass ratio is 2:1 i.e box A has twice the mass of B
Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:
Rate factor in the presence of catalyst:
Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:
![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;
Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
To delineate the the nature of the bonds that would be formed between the two elements, let us first write the electronic configuration of the two species;
Be = 2, 2
F = 2, 7
Beryllium is a metal with two valence electrons whereas fluorine is a halogen with seven valence electrons.
When Be loses two electrons it becomes isoelectronic with He;
Be → Be²⁺ + 2e⁻
Also, when fluorine gains an electron, it becomes isoelectronic with Ne;
F + e⁻ → F⁻
This loss and gain of electrons between the two elements creates an electrostatic attraction them and they enter into an electrovalent bond.
Hence;
Be²⁺ + 2F⁻ → BeF₂
