Answer:
e. When the indicator changes color and remains, 2 more drops of the solution in the buret should be added.
Explanation:
<em>Which statement is incorrect?
</em>
<em>Select one:
</em>
<em>
a. The volume of solution in the buret should be recorded when the endpoint is reached.
</em> CORRECT. At the endpoint, the titrant is expected to have completely reacted with the titrand.
<em>b. It is good practice to place the flask over a white background to better observe the color change of the solution.</em> CORRECT. Using a white background makes it easier to detect the changes in the color of the indicator.
<em>c. The endpoint is when the indicator changes color.</em>
CORRECT. This is the function of an indicator.
<em>d. Phenolphthalein will go from a colorless solution to a pink solution for the addition of NaOH to an acid solution.</em> CORRECT. Phenolphthalein is colorless when pH < 8.3 and it is pink when pH > 10.0.
<em>e. When the indicator changes color and remains, 2 more drops of the solution in the buret should be added.</em> INCORRECT. When the endpoint is reached, no more titrant should be added.
<span>(0.1875 moles)(98.004 g/mole) = 18.37575 g </span>
<span>In correct number of significant figures: 18.4 </span>
The most common way of preventing bumping is by adding one or two boiling chips to the reaction vessel. However, these alone may not prevent bumping and for this reason it is advisable to boil liquids in a boiling tube, a boiling flask, or an Erlenmeyer flask.
Answer:
255.51cm3
Explanation:
Data obtained from the question include:
V1 (initial volume) =?
T1 (initial temperature) = 50°C = 50 + 273 = 323K
T2 (final temperature) = - 5°C = - 5 + 237 = 268K
V2 (final volume) = 212cm3
Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:
V1/T1 = V2/T2
V1/323 = 212/268
Cross multiply to express in linear form
V1 x 268 = 323 x 212
Divide both side by 268
V1 = (323 x 212)/268
V1 = 255.51cm3
Therefore, the initial volume of the gas is 255.51cm3
The pH = 2.41
<h3>Further explanation</h3>
Given
5.0% by mass solution of acetic acid
the density of white vinegar is 1.007 g/cm3
Required
pH
Solution
Molarity of solution :
Ka for acetic acid = 1.8 x 10⁻⁵
[H⁺] for weak acid :
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Input the value :
![\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.839%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D0.00388%3D3.88%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~3.88%3D2.41)
