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3 years ago

Sand dollars typically live in the intertidal zone. Which adaptation do sand dollars most likely have?

Chemistry

2 answers:

AveGali [126]3 years ago

5 0

Answer:

burrowing in sandy or muddy substrates.

Explanation:

sand dollar also known as a sea cookie or snapper biscuit in New Zealand, or pansy shell in South Africa refers to species of extremely flattened, burrowing sea urchins belonging to the order Clypeasteroida. Some species within the order, not quite as flat, are known as sea biscuits.Sand dollars can also be called "sand cakes" or "cake urchins".

Sand dollars live beyond the low water line on top of or beneath the surface of sandy or muddy areas. Sand dollars are frequently found together on the ocean bottom.

ziro4ka [17]3 years ago

3 0

Neck to the sand in the water hope this help can i get brinlist??

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3 years ago

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2 years ago

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.