0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.
<h3>What is dilution?</h3>
Dilution is a process of making a solution of lower concentration from a solution of higher concentration by the addition of solvent to a given volume of the solution of higher concentration.
Dilution of solutions is done using the dilution formula in order to determine the given volume of diluent or stock solution required. The dilution formula is given below:
where:
- C1 = Initial concentration of enzyme
- C2 = Final concentration of enzyme
- V1 = Initial volume
- V2 = Final volume
For the enzyme dilution;
C1 = 1 mg/mL
C2 = 1/50 mg/ml = 0.02 mg/ml
V= ?
V2 = 5 ml
V1 = C2V2/C1
V1 = 0.02 * 5/1 = 0.1 mL
Therefore, 0.1 mL of the stock solution of the enzyme is taken and made up to 5.0 mL with 0.001M HCl in order to prepare a 50-fold diluted enzyme.
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Answer:
B. Salt, NaCl, is produced by the process of evaporation of seawater or brine. If the surface area of the water is increased, the same volume of water evaporates faster.
C. The Haber process combines hydrogen and nitrogen to make ammonia. The two gases are passed through a reactor under pressure and at high temperatures. If iron is added to the reactor, the yield of ammonia increases.
Explanation:
Evaporation of water is responsible for the production of sodium chloride also known as table salt. Sodium and chlorine are present in water. When more evaporation of water occurs, sodium and chlorine come close together forming sodium chloride. Haber process is responsible for the production of ammonia which is used as fertilizer. For speed up the process, catalyst is used such as iron in order to complete the reaction in less time. Iron binds hydrogen and nitrogen with each other.
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
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Answer:
The ionization equation is
⇄
(1)
Explanation:
The ionization equation is
⇄ (1)
As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.
The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,
![Ka = \frac{[HPO_{4}^{-2}] [H_{3}O^{+}]}{[H_{2}PO_{4}^{-}] [H_{2}O]} = 6.2x10^{-8}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BHPO_%7B4%7D%5E%7B-2%7D%5D%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BH_%7B2%7DPO_%7B4%7D%5E%7B-%7D%5D%20%5BH_%7B2%7DO%5D%7D%20%3D%206.2x10%5E%7B-8%7D)
The pKa is
The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.
In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.
If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.
The oxidation number of P in Mg3P2 is 3. When writing ionic compounds, you swap the oxidation numbers and add them as the subscript.
