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AlladinOne [14]
3 years ago
8

Which statements about the operation of an automated micropipettor are true? The volume dial has already been set, do not alter

it. The volume dial has not been set, adjust it to the desired setting. A new disposable plastic tip is required for each sample. The point of initial resistance is the level of depression that will fill the pipette with the desired volume of solution.
Chemistry
1 answer:
yanalaym [24]3 years ago
4 0

Answer:

The correct statement is that the point of initial resistance is the level of depression that will fill the pipette with the desired volume of solution.

Explanation:

The pipette can be otherwise stored vertically or horizontally, this has nothing to do with the operation, and before the operation, the dial needs to be set. Also, the plastic pipette tip should not be ejected at all, thus, no new disposable plastic tip is required for each sample.

The correct statement is that the point of initial resistance is the level of depression that will fill the pipette with the desired volume of solution.

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What is the molarity of a solution that is prepared by adding 57.1 g of toluene (c 7 ​ h 8 ​ ) (density = 0.867 g/ml) to a 250 m
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Answer:

2.48 mol/L.

Explanation:

  • The molarity of the solution can be expressed as <em>the number of moles of solute in 1.0 liter of the solution, </em>(M = n / V).
  • It is also can be calculated from the relation:

<em>M = (mass / molar mass) solute x (1000 / V of solution)</em>

The solute is toluene and the solvent is benzene.

mass of toluene (solute) = 57.1 g,

molar mass of toluene (solute) = 92.14 g/mol.

volume of the solution = 250 ml.

∴ M =  (mass / molar mass) solute x (1000 / V of solution) = [(57.1 g / 92.14 g/mol) x (1000 / 250 ml)] = 2.48 mol/L.

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3 years ago
Which layer of Earth is responsible for the movement of tectonic plates?
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2 years ago
determine the average volume from the following volume reading: 3.00ml, 2.0ml, 2.987 x 10^-3 l, and 3.4856 ml
lesya [120]

Answer:

The average of given values is 2.1221 ml

Explanation:

Given data:

Given measurements = 3.00 ml , 2.0 ml, 2.987 × 10⁻³ml , 3.4856 ml

Average value of given measurements = ?

Solution:

Formula:

Average value = sum of all measurement  / total number of measurements

2.987 × 10⁻³ml = 0.002987 ml

Now we will put the  values.

Average value = 3.00 ml + 2.0 ml + 0.002987 ml+ 3.4856 ml / 4

Average value = 8.488587 ml / 4

Average value =  2.1221 ml

The average of given values is 2.1221 ml

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