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JulijaS [17]
3 years ago
10

Determine whether each practice is recommended to keep a micropipette clean and functioning properly.

Chemistry
1 answer:
Natasha2012 [34]3 years ago
8 0

Answer:

1. A

2. B

3. B

4. A

5. B

Explanation:

1. Recommended. That way cross-contamination is prevented.

2. Not recommended. The inner mechanism of the micropipette should not come in contact with any liquid whatsoever.

3. Not recommended. It contradicts statement 1.

4. Recommended. That way the micropipette remains completely vertical, minimizing the risks of being decalibrated.

5. Not recommended. If it lays on the bench it will be horizontal, thus contradicting statement 4.

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A 1.10 g sample contains only glucose and sucrose. When the sample is dissolved in water to a total solution volume of 25.0L, th
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Answer:

\large \boxed{79 \, \%}

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then  

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g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

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1. Set up the osmotic pressure condition

Π = cRT, so

\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4)  & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}

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