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vazorg [7]
3 years ago
13

For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to

label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won't this experiment work?
A. There is no radioactive isotope of nitrogen.B. Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long.C. Meselson and Stahl already did this experiment.D. Although there are more nitrogens in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are more radioactive.E. Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.
Chemistry
1 answer:
julia-pushkina [17]3 years ago
4 0

Answer:

E

Explanation:

E. Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.

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What is the temperature of a gas at STP?
Savatey [412]

Answer:

273 K

Explanation:

The temperature of a gas at STP is 273 K. This is equal to 0°C or 32°F.

3 0
3 years ago
if the temperature of this equilibrium was increased, what would happen to the equilibrium yield of chlorine?
NARA [144]

Answer:

the temperature of the system decreases

3 0
2 years ago
Calculate the enthalpies of formation, ΔHf∘, of the group 1 fluoride compounds from their elements using the Born–Haber cycle.
maxonik [38]

Answer:

Enthalpy of formation of KF = -555 kJ/mol

Enthalpy of formation of CsF = -539kJ/mol

Explanation:

Explanations are provided in the attachment below

5 0
4 years ago
A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg
Brut [27]

Answer:  1.0\times 10^2g

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

K_f = freezing point constant = 3.96^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}

Weight of solvent (X)= 950 g = 0.95 kg  

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}

x=1.0\times 10^2g

Thus 1.0\times 10^2g urea was dissolved.

8 0
4 years ago
Convert 123 in scientific notation
dalvyx [7]
✡ Answer: 1.23*10^2 ✡


- - Add a decimal at the end (to the right) and count till you get to the first number.
So now you have 1.23

- - Now you always want to times it by 10 to the power of how many times you moved it over, in this case, 2

Final answer: 1.23*10^2

✡Hope this helps✡


4 0
3 years ago
Read 2 more answers
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