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o-na [289]
3 years ago
5

1. After clamping a buret to a ring stand, you notice that the set-up is tippy and unstable. What should you do to stabilize the

set-up?
A. Move the buret clamp to a ring stand with a larger base.
B. Hold the buret in place by hand during the titration.
C. Use a different size buret that is more stable in the set-up
D. Clamp another buret to the ring stand to balance out the set-up.
Chemistry
1 answer:
vampirchik [111]3 years ago
7 0

Answer:

A. Move the buret clamp to a ring stand with a larger base.

Explanation:

The ring stands are used to hold burettes, light in weight to avoid loss of stability, that is why it is necessary to change the size of the ring stand so that it can support the buret that we are going to use.  It is not recommended to balance it with the hand since it would give us an inaccurate result in the titration.

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Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissol
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Here is the complete question.

Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissolves readily in water and polar organic solvents like ethanol. Calculate the mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol (CH3CH2OH; density = 0.7893 g/mol). Round to four significant digits

Answer:

0.9567 mol

Explanation:

Given that:

mass of glycerol = 1.61 g

molar mass of glycerol = 92.1 g/mol

no of mole = \frac{mass}{molar mass}

∴ number of moles of glycerol (n_{glycerol}) = \frac{1.61}{92.1}

= 0.0175 mol

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Density of ethanol = 0.7893 g/mL

Since Density = \frac{mass}{volume}

∴  mass of ethanol = density of ethanol × volume of ethanol

mass of ethanol =  0.7893 g/mL × 22.60 mL

mass of ethanol =  17.838 g

Number of moles of ethanol (n_{ethanol}) = \frac{17.838}{46.0684}

= 0.387 mole

∴ the mole fraction of the solvent can be determined as:

X_{solvent} = X_{ethanol} = \frac{n_{ethanol}}{n_{glycerol}+n_{ethanol}}

=\frac{0.387 mol}{(0.0175+0.387)mol}

= \frac{0.387mol}{0.4045mol}

= 0.95673671199

≅ 0.9567 mol

∴ The mole fraction of the solvent in a solution that contains 1.61 g glycerol dissolved in 22.60 mL ethanol is = 0.9567 mol

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