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ladessa [460]
2 years ago
13

Which group of archaea are chemoautotrophic anaerobes that use hydrogen as an electron donor and sulfur compounds as terminal el

ectron acceptors?
Chemistry
1 answer:
vodomira [7]2 years ago
7 0

The group of archaea that are chemoautotrophic anaerobes that use hydrogen as an electron donor and sulfur compounds as terminal electron acceptors are the

Thermoacidophiles.

<h3>What are chemoautotrophic anaerobes?</h3>

The chemoautotrophic anaerobes are those organisms that derive energy for survival through the reduction of the following compounds:

  • CO2 to CH4, or

  • SO4 to H2S.

Thermoacidophiles is a typical example of the group of archaea that are chemoautotrophic anaerobes that use hydrogen as an electron donor and sulfur compounds as terminal electron acceptors.

Learn more about anaerobes here:

brainly.com/question/16188326

#SPJ1

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A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

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4 years ago
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How many grams of material is lost in the aqueous phase if two extractions are carried out on 100 mL of a 5% (m/v) aqueous solut
san4es73 [151]

Answer:

4.94g of material

Explanation:

Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:

Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>

100mL of a 5% solution contains 5g of material in 100mL of water. Thus:

8 = X / 100mL / (5g-X) / 100mL

<em>Where X is the amount of material in grams that comes to the organic phase.</em>

8 = X / 100mL / (5g-X) / 100mL

8 = 100X / (500-100X)

4000 - 800X = 100X

4000 = 900X

4.44g = X

<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>

And will remain 5g-4.44g = 0.56g.

In the second extraction:

8 = X / 100mL / (0.56g-X) / 100mL

8 = 100X / (56-100X)

448 - 800X = 100X

448 = 900X

0.50g = X

<em>In the second extraction, you will extract 0.50g of material</em>

Thus, after the two extraction you will lost:

4.44g + 0.50g = <em>4.94g of material</em>

<em></em>

6 0
3 years ago
Animal and plant cells all have the following organelles
pav-90 [236]

Answer: C

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3 years ago
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