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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

You might be interested in
Chem pls help and look at pics
ser-zykov [4K]

Answer:

1. 16 protons, 18 electrons, 16 neutrons

2. Ba (2+) + 2 NO3 (-) + 2 Na (+) + SO4 (-) = 2 Na (+) + 2 NO3 (-) + BASO4

Explanation:

1. By looking at the number in the top left corner, we know the mass is 32. And by looking at the number in the bottom left, we know that we have 16 protons. Now the way to find neutrons is to subtract the number of protons from the mass (32-16) which will give us 16 neutrons. Now the electrons are a bit tricky. Usually, atoms have the same number of protons and electrons, hence why atoms don't usually have a charge. However, in this case we have a charge of 2- (as seen in the top right corner) so we need to add two to the number of protons to find our number of electrons (electrons are negatively charged, so if there are more of them the atom will have an overall negative charge) and when we do that we get 18. So the final answer is 16 protons, 18 electrons, and 16 neutrons.
2. This one I'm not 100% sure on since the pictures are kinda wonky but essentially the way net ionic equations are written is that whatever is aqueous/soluble (aq) will be written separate from whatever it was bonded to with the charge shown, but solids (s), liquids (l), and gases (g) are written still bonded without the charge being shown. So in this case, everything in the reaction is aqueous except for the BaSO4 product. The parentheses are used to show charges, but won't be written in an actual equation, and the equal sign is the same as the arrow. So you're equation is written as Ba (+) + 2 NO3 (-) + 2 Na (+) + So4 (2-) = 2 Na (+) + NO3 (-) + BaSO4.
Hope this helps! :)

6 0
2 years ago
The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D
Westkost [7]

Answer:

a) K_{2} S and NH_{4} Cl :

There are no insoluble precipitate forms.

b) Ca Cl_{2} and (NH_{4} )_{2} Co_{3} :

There are the insoluble precipitates of CaCo_{3}  forms.

c) Li_{2}S and MnBr_{2} :

There are the insoluble precipitates of MnS  forms.

d) Ba(No_{3} )_{2} and Ag_{2} So_{4} :                        

As Ag_{2} So_{4} is insoluble, therefore no precipitate forms.

e) Rb_{2}Co_{3} and NaCl:

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- K_{2} S ⇒ soluble, NH_{4} Cl ⇒ soluble.

                                          KCl ⇒ soluble, (NH_{4})_{2} S  ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- Ca Cl_{2} ⇒ soluble, (NH_{4} )_{2} Co_{3} ⇒ soluble.

                                        CaCo_{3} ⇒ insoluble, NH_{4} Cl  ⇒ soluble.

There are the insoluble precipitates of CaCo_{3}  forms.

c)

Solubility rule suggests:- Li_{2}S ⇒ soluble, MnBr_{2} ⇒ soluble.

                                        LiBr ⇒ soluble, MnS  ⇒ insoluble.

There are the insoluble precipitates of MnS  forms.

d)

Solubility rule suggests:- Ba(No_{3} )_{2} ⇒ soluble, Ag_{2} So_{4} ⇒insoluble.

                                     

As Ag_{2} So_{4} is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- Rb_{2}Co_{3} ⇒ soluble, NaCl ⇒ soluble.

                                        RbCl ⇒ soluble, Na_{2} Co_{3}  ⇒ soluble.

There are no insoluble precipitates forms.

6 0
3 years ago
Cis-platin is a chlorine-containing chemotherapy agent with the formula pt(nh3)2cl2. What is the mass of one mole of cis-platin?
marishachu [46]

Cis-platin is a chemotherapy agent used to treat and kill cancerous cells in patients. One mole of cis-platin has a mass of 300.06 grams/mol. Thus, option B is correct.

<h3>What is a chemotherapy agent?</h3>

A chemotherapy agent is an alkylating agent that is used to treat cancer and is given to reduce the infection or to relieve the symptoms. Cis-platin (Pt(NH₃)₂Cl₂) is one of the chemotherapy agents that treat lung, ovarian, and neck cancer, etc.

The mass of one mole of Cis-platin is calculated as,

Moles = mass ÷ molar mass

Where,

Moles = 1 mole

The molar mass of (Pt(NH₃)₂Cl₂) is calculated as,

195.06 g/mole + 2(17g/mole) + (35.5)2 = 300.06 grams

Substituting values to calculate mass:

Mass = Molar mass × moles

= 300. 06  × 1

= 300.06 grams/mol

Therefore, option B. 300.06 gm/mol is the mass of Cis-platin.

Learn more about chemotherapy agents here:

brainly.com/question/14260402

#SPJ4

Your question is incomplete, but most probably your full question was, Cis-platin is a chlorine-containing chemotherapy agent with the formula pt(nh3)2cl2. What is the mass of one mole of cis-platin?

  • 488.91 g/mol
  • 300.06 g/mol
  • 492.37 g/mol
  • 283.02 g/mol
  • 2860.5 g/mol
8 0
2 years ago
An aqueous solution of licl is 34.0 % licl. what mass of water is present in 250.0 g of the solution? (density of water is 1.00
adell [148]
<span>If the aqueous solution is 34% Licl then it is 100 - 34% water = 66% From the calculation we've found out that it is 66% water. Then we need to find the weight from a 250 g solution. 66/100 * 250 = 165g Hence it is 165g</span>
4 0
3 years ago
Read 2 more answers
Which is an example of a solution?
Rudiy27
The answer is C. Salt and water is a solution 
3 0
3 years ago
Read 2 more answers
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