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valina [46]
2 years ago
7

Describe how the meniscus is used to read volume measurements on a graduated cylinder

Chemistry
2 answers:
LekaFEV [45]2 years ago
5 0
The liquid in a graduated cylinder curves up at the edge where the liquid meets the wall of the cylinder. This curve in the liquid is called the meniscus and is used to determine the volume of liquid in a graduated cylinder by observing the measurement tick closest to the bottom of the meniscus.
katrin2010 [14]2 years ago
4 0

Answer:Tthe liquid in a graduated cylinder curves up at the edge where the liquid meets the wall of the cylinder. This curve in the liquid is called the meniscus and is used to determiine the volume of liquid in a graduated cylinder by observing the measurement tick closest to the bottom of the meniscus.

Explanation:

Hope it helps :D

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How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride (FM 121.135) to give a pH of 7.60 in a fina
liubo4ka [24]

Answer:

41.64mL of NaOH 0.500M must be added to obtain the desire pH

Explanation:

It is possible to find pH of a buffer by using H-H equation, thus:

pH = pka + log [A⁻] / [HA]

<em>Where [HA] is concentration of the weak acid TRIS-HCl and [A⁻] is concentration of its conjugate acid.</em>

Replacing in H-H equation:

7.60 = 8.072 + log [A⁻] / [HA]

0.3373 =  [A⁻] / [HA] <em>(1)</em>

10.0g of TRIS-HCl (Molar mass: 121.135g/mol) are:

10.0g ₓ (1mol / 121.135g) = 0.08255 moles of acid. That means moles of both the acid and conjugate base are:

[A⁻] + [HA] = 0.08255 <em>(2)</em>

Replacing (1) in (2):

0.3373 =  0.08255 - [HA] / [HA]

0.3373[HA] =  0.08255 - [HA]

1.3373[HA] = 0.08255

<em>[HA] = 0.06173 moles</em>

Thus:

[A⁻]  = 0.08255 - 0.06173 = 0.02082 moles [A⁻]

The moles of A⁻ comes from the reaction of the weak acid with NaOH, that is:

HA + NaOH → A⁻ + H₂O + K⁺

Thus, <em>you need to add 0.02082 moles of NaOH to produce 0.02082 moles of A⁻. </em>As NaOH solution is 0.500M:

0.02082 moles NaOH ₓ (1L / 0.500mol) = 0.04164L of NaOH 0.500M =

<h3>41.64mL of NaOH 0.500M must be added to obtain the desire pH</h3>

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