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stepan [7]
3 years ago
12

A researcher wants to determine if a unicellular organism he discovered is an autotroph or a heterotroph. He radioactively label

s the carbon in CO2 and C6H12O6, and exposes one culture of his organism to the labeled CO2 and another culture to the labeled C6H12O6. What would happen if his organism is an autotroph?
A. Labeled carbon would be seen in the carbohydrates of organisms exposed to CO2.
B. Labeled carbon would be seen in the carbohydrates of organisms exposed to C6H12O6.
C. Labeled carbon would not be seen in the carbohydrates of either culture.
D. Labeled carbon would be seen in carbohydrates of both cultures.
Chemistry
1 answer:
GaryK [48]3 years ago
4 0

Answer:

A

Explanation:

Autotrophs utilize the energy from  sunlight to reduce carbon dioxide to carbohydrates (glucose). The energy from the sunlight is used to split water into H+ and O2- and the H+ used in the reduction process. The labeled carbon in the carbon dioxide will, therefore, be incorporated by the autotrophs in the carbohydrates made in photosynthesis.  

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g n the Ideal Gas Law lab, how is the temperature of the hydrogen gas determined? Select one: The pressure of the gas is determi
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Answer:

The volume of the gas is determined, which will allow you to calculate the temperature.

Explanation:

According to Charles law; the volume of a given mass of an ideal gas is directly proportional to its temperature at constant pressure.

This implies that, when the volume of an ideal gas is measured at constant pressure, the temperature of the ideal gas can be calculated from it according to Charles law.

Hence in the Ideal Gas Law lab, the temperature of an ideal gas is measured by determining the volume of the ideal gas.

4 0
2 years ago
A black hole can be considered a star that has
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7 0
3 years ago
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
The isomerization reaction, ch3nc → ch3cn, is first order and the rate constant is equal to 0.46 s-1 at 600 k. what is the conce
Harlamova29_29 [7]
According to this formula :
㏑[A] /[Ao] = - Kt 
when we have Ao = 0.3 m 
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3
6 0
3 years ago
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