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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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Answer:

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4 0
3 years ago
A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug
OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 =  log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] <em>(1)</em>

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M <em>(2)</em>

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = <em>0.035M</em>

Thus: [CH₃COO⁻] = 0.100M - 0.035M = <em>0.065M</em>

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = <em>7.429x10⁻³ moles of CH₃COO⁻</em>

<em />

Moles of CH₃COOH  before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles +  2.646x10⁻³ moles = <em>8.071x10⁻³ moles of CH₃COOH</em>. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = <em>0.296</em>

4 0
3 years ago
What does the stonefish eat
solmaris [256]
Croat or stones stones
5 0
3 years ago
At 273 k and 1.00 x 10^-2 atm, the density of a gas is 1.24 x 10^-5 g/cm3.
bija089 [108]

Root mean square velocity is the square root of the mean of the squares of speeds of different molecules. From kinetic theory of gas, the formula of root mean square velocity=C_{rms}= √\frac{3RT}{M}=√\frac{3PV}{M}=√\frac{3P}{d}, where, R= Universal gas constant, T= Absolute temperature, P= Pressure, V= Volume of gas, d= Density of gas.

Given,  T=273 K, P=1.00 x 10⁻² atm, d=1.24 x 10⁻⁵ g/cm³.

(a) Using the formula C_{rms}=√\frac{3P}{d}=√(3X1.00X10⁻²)/(1.24X10⁻⁵)=49.18

(b) Molar mass can be determined by using the formula C_{rms}=√{3RT}{M}

49.18=√\frac{3X8.314X273}{M}

49.18²=√(3X8.314X273)/M

M=\frac{3X8.314X273}{49.18^{2} }

M=1.67 ≅ 2

Molecular mass is 2.

(c) The gas is Helium (He) whose molecular mass is 2.

8 0
3 years ago
Read 2 more answers
For the reaction below, complete the rate expression that relates the change in concentration with respect to time to the rate o
Ann [662]

Answer: Rate in terms of disappearance of NO = -\frac{1d[NO]}{2dt}

Rate in terms of disappearance of Cl_2= -\frac{1d[Cl_2]}{1dt}

Rate in terms of appearance of NOCl = \frac{1d[NOCl]}{2dt}

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

2NO+Cl_2\rightarrow 2NOCl

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of  = -\frac{1d[NO]}{2dt}

Rate in terms of disappearance of = -\frac{1d[Cl_2]}{1dt}

Rate in terms of appearance of NOCl = +\frac{1d[NOCl]}{2dt}

5 0
3 years ago
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