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3 years ago

_____ helps to destroy animals and plants when they die, preventing them from fossilizing.

Chemistry

2 answers:

Bogdan [553]3 years ago

8 0

Answer:

Decomposers

Explanation:

Decomposers are defined as those organisms that decomposes the organic components of the dead organisms. For example, microbes and saprophytes.

When plants and animals dies, the decomposers eats up the organic substance. These consumption of organic substances helps in the growth and expansion of the microbes.

The decomposers does not allow the organisms to undergo fossilization process. It prevents them from turning into a fossil. The organism completely decays if there is no hard parts present in it.

Kazeer [188]3 years ago

5 0

It think it would be oxygen because it would then make what ever died begin to decompose.

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3 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

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3 years ago

24-Complete the following sentence:

docker41 [41]

Answer:

B. Solvent

Explanation:

In osmosis, water always moves from low solute concentration to high solute concentration. SOLUTE NEVER MOVES AS IT CANNOT PASS THE SELECTIVELY PERMEABLE MEMBRANE. alot of caps but need to stress this concept cuz otherwise this concept gets very confusing

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3 years ago