Sodium thiosulfate is used as a reducing agent in the Kjeldahl procedure. It is used to prevent the interference of certain substances when determining nitrogen content in organic compounds.
During the digestion step of the Kjeldahl procedure, the organic sample is digested with sulfuric acid. This breaks down the organic compounds and converts the nitrogen present into ammonium sulfate. Some models may contain substances that can interfere with the next analysis. This can lead to inaccuracies in the outcomes.
Using sodium thiosulfate in the modified Kjeldahl procedure minimizes interference from certain compounds. The accuracy of the nitrogen determination is improved. This helps us to ensure that the results obtained are correct and instance of the actual nitrogen content in the sample.
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The role of Sodium thiosulfate in the modified Kjeldahl procedure is to act as a reducing agent and prevent interference from certain substances, such as nitrites and oxidizing agents, which can affect the accuracy of the nitrogen content determination in organic compounds.
In the modified Kjeldahl procedure, Sodium thiosulfate (Na2S2O3) plays a crucial role as a reducing agent. This procedure is commonly used for the determination of nitrogen content in organic compounds. The addition of Sodium thiosulfate to the reaction mixture serves to prevent interference from certain substances that can react with the reducing agent used in the procedure.
When performing the modified Kjeldahl procedure, it is important to accurately determine the nitrogen content in the sample. However, certain substances, such as nitrites and other oxidizing agents, can interfere with the reaction and affect the accuracy of the results. These interfering substances can react with the reducing agent, leading to inaccurate measurements.
By adding Sodium thiosulfate to the reaction mixture, these interfering substances are neutralized. Sodium thiosulfate acts as a reducing agent, effectively preventing the interference of nitrites and other oxidizing agents. This allows for a more accurate determination of the nitrogen content in the sample.
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Answer:
Mass of water (m) = 25 grams = 0.025 kg (since 1 g = 0.001 kg)
Specific heat of water (c) = 4.18 J/(g°C) = 4.18 J/(kg°C)
Initial temperature () = 22°C
Final temperature ( )= 45°C
Change in temperature (ΔT):
ΔT=-=45°−22°=23°
Now, calculate the heat energy (Q)
Q=mass×specific heat×ΔT
Q=0.025kg×4.18J/(kg°C)×23°C
Q≈2.44kJ
So, the heat energy for this scenario is approximately 2.44 kilojoules (kJ).
B. Calcium chloride (CaCl2)
C. Potassium Iodide (KI)
D. Glucose (C6H12O6)
Answer:
Explanation:
The colligative properties depend on the concentration of particles (molecules or ions) of solute dissolved and not on the identity of the solute.
For your reference the colligative properties are: i) boiling point increase, ii) freezing point depression, iii) vapor pressure lowering, and iv) osmotic pressure.
When the solute is a ionic compound the ionization yields several ions, so for ionic solutes you must predict first the number or ions formed per unit of compound and then predict the effect on the colligative properties: the greater the number of ions per unit of compound the greater the effect on colligative properties.
So, just set the ionization equations for each compound:
Then, since calcium chloride, CaCl₂, produces the greater number of ions it is the solute that productes the greatest effect on colligative properties.
Answer:
D. Calcium chloride (CaCl2)
Explanation:
Founder's Education/ Educere Answer
Answer:
The final solution will be clear and colorless.
Explanation:
The balanced reaction equation is:
NaOH + HCl ⇒ H₂O + NaCl
The amount of NaOH that is added is calculated as follows:
(20.0mL)(1.0mol/L) = 20 mmol NaOH
Similarly, the amount of HCl that is added is as follows:
(10.0mL)(1.0mol/L) = 10 mmol HCl
Since HCl and NaOH react in 1:1 proportions, the HCl is the limiting reaction. 10 mmmol of HCl will neutralize 10 mmol of NaOH, leaving 10 mmol of NaOH. The volume of the mixed solution is 30.0 mL, so the concentration of NaOH in the final solution is:
(10 mmol)/(30.0mL) = 0.3333 M NaOH
The pOH of the final solution is:
pOH = -log([OH⁻) = -log(0.3333) = 0.477
The pOH is related to the pH as follows:
pH = 14 - pOH = 14 - 0.477 = 13.5
At a pH of 13.5, phenolphthalein is colorless. The final solution will be observed as clear and colorless.
High temperatures increase the activation energy of the reaction.
High temperatures make the gas molecules move more quickly.
The reaction becomes exothermic at high temperatures.
this is wrong do not put this if your with FLVS
the correct answer is C i got 100% on it