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3 years ago

Nitrifiers are bacteria

Chemistry

1 answer:

Delicious77 [7]3 years ago

5 0

Answer:

(D.) Nitrifiers are bacteria that generate nitrites or nitrates.

Explanation:

In the nitrogen cycle which occurs in nature, ammonia and ammonium compounds in the soil from organic sources and are converted to nitrites and nitrates by aerobic microorganisms.

Nitrifiers, as the name implies, are these such aerobic bacteria which oxidize inorganic constituents in the soil to generate energy. Examples of these nitrifiers are nitrobacter and nitrosomonas.

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A certain metal M forms a soluble sulfate salt M2SO4. Suppose the left half cell of a galvanic cell apparatus is filled with a 5

quester [9]

Answer:

Explanation:

For a certain metal M by which the electrodes and the solution M_2SO_4 the cell notation is:

SALT BRIDGE

ANODE ↓ CATHODE

$M(s)|M_2SO_4_{(aq)} (50.0 \ mM)\Big {|} \Big {|} M_2SO_4_{(aq)} (5.00 \ M ) |M(s)$

Since the standard electrode potential is less positive on the left side, a negative anode electrode is used, and oxidation occurs at the anode.

$i.e \ \ M(s) \to M^{2+} _{(aq)} + 2 e^- \ \ \ (oxidation)$

The right side of the cell cathode is placed which is positive in sign and aids in reduction since the normal reduction potential is less negative.

$i.e \ \ M^{2+} _{(aq)} + 2 e^- \to M(s) \ \ \ (reduction)$

As a result, the correct answer for the positive electrode is the right side.

$M_2SO_4 \to M^{2+} + SO_4^{2-}$

Using Nernst Equation:

$E_{cell} = - \dfrac{RT}{nF }log ( \dfrac{reduction \ half }{oxidation \ half})$

$E_{cell} = - \dfrac{2.303 \times 8.314 \times 293}{2 \times 96485 }log ( \dfrac{5.0 0 }{50 \times 10^{-3}})$

$E_{cell} = -0.058 \ V$

5 0

3 years ago

One way to account for the mass "lost" in a reaction that involves a gas would be to________.

sergiy2304 [10]

Answer:

Place the experiments on a scale

Explanation:

If you place two of the same experiments on a scale, and cover one with a cup, the one with the cup will "lose" mass, while the other won't, due to the cup being a "containment field" I guess you could call it.

5 0

3 years ago

Question 10

creativ13 [48]

Answer:

M₂ = 0.08 M

Explanation:

Given data:

Initial volume = 25.0 mL

Initial molarity = 0.500 M

Final volume = 150.0 mL

Final molarity = ?