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lidiya [134]
3 years ago
13

The area of the gold electrodes on the quartz crystal microbalance at the opening of Chapter 2 is 3.3 mm^2. One gold electrode i

s covered with DNA at a surface density of 1.2 pmol/cm2. (a) How much mass of the nucleotide cytosine (C) is bound to the surface of the electrode when each bound DNA is elongated by one unit of C
Chemistry
1 answer:
aleksandrvk [35]3 years ago
5 0

This question is incomplete, the complete question is;

The area of the gold electrodes on the quartz crystal microbalance at the opening of Chapter 2 is 3.3 mm^2. One gold electrode is covered with DNA at a surface density of 1.2 pmol/cm2.

(a) How much mass of the nucleotide cytosine (C) is bound to the surface of the electrode when each bound DNA is elongated by one unit of C. The mass formula mass of the bound nucleotide is cytosine + deoxyribose + phosphate = C9H10N3O6P = 287.2 g/mol

Answer: mass of the nucleotide (c) bound is 11.37 g

Explanation:

Given that the area of gold electrodes = 3.3 mm^2

surface density of one gold electrode = 1.2 pmol/cm^2

that is to that in every 1 cm^2 of area, 1.2 pmol DNA is present

therefore

mass of nucleotide present in 3.3 mm^2 is;

= (1.2/100 * 3.3) pmol

= 0.0396 pmol

we were given that formula mass of the bound nucleotide = 287.2 g/mol

so

mass of the nucleotide (c) bound = ( 287.2 * 0.0396 )g

mass of the nucleotide (c) bound =  11.37 g

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Answer:

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

Explanation:

We will first write the balanced equation for this scenario

3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2

3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2

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Using the mole ratio of 3:2, convert each to moles of sodium phosphate.

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22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

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The answer to your question is V2 = 434.7 l

Explanation:

Data

Volume 1 = V1 = 240 l                             Volume 2 = ?

Temperature 1 = T1 = 479°K                   Temperature 2 = T2 = 293°K

Pressure 1 = P1 = 300 KPa                      Pressure 2 = P2 = 101.325 Kpa

Process

1.- Use the combined gas law to solve this problem

                P1V1/T1 = P2V2/t2

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               V2 = P1V1T2 / T1P2

2.- Substitution

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3.- Simplification

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