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
Answer:
1.40*10⁻² M
Explanation:
We have the solubility formula
Solubility,
S = KH*P
where
KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm
P = atmospheric pressure = 0.400 atm
Hence, we have
S = KH*P
= (3.50*10⁻² mol/L.atm)*(0.400 atm)
= 1.40*10⁻² mol/L
But 1 mol/L = 1 M,
Hence, the answer (1.40*10⁻² mol/L
) is equivalent to
= 1.40*10⁻² M
Answer:
Ca
2+
<K + <Ar<Cl − <S 2−
Explanation:
Ar,K +
,Cl −
,S 2−
,Ca 2+
have the same number of electrons. Their radii would be different because of their different nuclear charges. The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have the larger radius. In this case, the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size. Hence the correct order will be Ca
2+ <K + <Ar<Cl − <S 2−
Answer:
mechanical energy to electrical energy to light energy
