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3 years ago

Bacteria and fungi break down organisms into nutrients. They are known as

1 answer:

Rudiy273 years ago

5 0

The right answer for the question that is being asked and shown above is that: "decomposers." Bacteria and fungi break down organisms into nutrients. They are known as decomposers. They are responsible of breaking down foods and nutrients.

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1. Why don't scientists stop with the first step in the Scientific Method, making observations? A. They don't trust their own ob

natulia [17]

Answer:

Explanation:

Scientists don't stop with the first step of their experiment because they want other scientists' opinions because they may not trust their own observations.

Scientists don't stop with the first step of their experiment because they would rather plan and run experiments than just observe the world around them

Hope any one of these helps you

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3 years ago

What would be the mass of a 40 cm3 piece of manganese?

Digiron [165]

1.738 g/cm<span>3 is the density of magnesium just multiply it you will get 69.52g.That's my final answer.</span>

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3 years ago

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3 years ago

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How many electrons, protons and neutrons does the element Argon have?

erik [133]

Answer:

18 Protons

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2 years ago

3. Balance each of the following redox reactions in the listed aqueous environment. (4 pts each, 8 pts total) Crs)NO3 (a) Cr (a)

GrogVix [38]

Explanation:

(a) The given reaction equation is as follows.

$Cr(s) + NO^{-}_{3}(aq) \rightarrow Cr^{3+}(aq) + NO(g)$ (acidic)

So, here the reduction and oxidation-half reactions will be as follows.

Oxidation-half reaction: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$

Reduction-half-reaction: $NO^{-}_{3} + 3e^{-}(aq) \rightarrow NO(g)$

As total charge present on reactant side is -1 and total charge present on product side is +3. And, since it is present in aqueous medium. Hence, we will balance the charge for this reaction equation as follows.

$Cr(s) + NO^{-}_{3}(aq) + 4H^{+}(aq) \rightarrow Cr^{3+}(aq) + NO(g) + 2H_{2}O(l)$ (acidic)

(b) The given reaction equation is as follows.

$HCO^{-}_{3}(aq) + Ag(s) + NH_{3}(aq) \rightarrow H_{2}CO(aq) + Ag(NH_{3})^{+}_{2}(aq)$ (basic)

So, here the reduction and oxidation-half reactions will be as follows.

Reduction-half reaction: $HCO^{-}_{3}(aq) + 4e^{-} \rightarrow H_{2}CO(aq)$

Oxidation-half reaction: $Ag(s) \rightarrow Ag(NH_{3})^{+}_{2}(aq) + 1e^{-}$

Hence, to balance the number of electrons in this equation we multiply it by 4 as follows.

$4Ag(s) \rightarrow 4Ag(NH_{3})^{+}_{2}(aq) + 4e^{-}$

Therefore, balancing the whole reaction equation in the basic medium as follows.

$H_{2}CO(aq) + 4Ag(NH_{3})^{+}_{2}(aq) + 5OH^{-}(aq) \rightarrow HCO^{-}_{3}(aq) + 4Ag(s) + 8NH_{3}(aq) + 3H_{2}O(l)$

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3 years ago