Its c ok im sorry if i got the last answer wrong
Answer: The correct option is ALL OF THE ABOVE.
Explanation:
TITRATION is a type of volumetric analysis which is used for determining the concentration of solutions. In this process a specific volume of a solution is placed in a conical flask by means of a pipette and small quantities of a second solution is slowly added from a burette until the end point is reached. This is determined by a means of an indicator which shows a characteristic colour change.
During titration, the following precautions should be followed to avoid errors and maintain standardisation in the experiment.
--> Any air bubble in the burette and pipette must be removed during measurement
--> the burette tap should be tightened to avoid leakage.
--> Remove the funnel from the burette before taking any reading to avoid errors in reading the volume.
--> use the base solution such as Sodium Hydroxide Solution to rinse the burette after washing with soap and tap water:
• to remove any air bubble and fill it's tip
• to remove any residual liquid from the water and soap solution which may interfere with the results of the experiment.
• to check if the burette is in good condition.
Therefore all of the above options to the question are correct.
This can be, for example, halogensubstituted hydrocarbons.
CCl₄, C₂F₆.
Or halides halocarboxylic acids, and other compounds.
O
II
Cl₃C-Cl
Answer:
compound
Explanation:
because now there are one together
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C