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Virty [35]
3 years ago
13

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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Based on the results you observed for the iodine test and Benedict’s test, is it better to detect enzyme activity by measuring t
dlinn [17]

Answer:

Is better use the Benedict's test by the increase in the amount of the products if the enzyme is a reductase

Explanation:

The Benedict's test works by the reaction of the reducing sugars with the ion cupric of the reactive. If the enzyme is a reductase (degrades polysaccharides into bi o monosaccharides), it should cut the polysaccharide bond and the products would react with the Benedict's cupric ion

I hope you undestand me

3 0
3 years ago
what conclusions can be made about the relationship between metallic character and the atomic radius?
kolezko [41]

We have to get the relationship between metallic character and atomic radius.

Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.

If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.

With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.

With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.

5 0
2 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
What is the element of H₂O
slega [8]

Answer:

It’s water bro

Explanation:

6 0
3 years ago
Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, ml, an
vfiekz [6]
First, since l = n-1,
5,4,-5,1/2 and 2,1,0,1/2 are the only answer choices left.
Next, since ml = -l to l,
2,1,0,1/2
is the answer because in 5,4,-5,1/2, the ml value of -5 is not in the range of -4 to 4, as notes by the value 4 for l.
8 0
2 years ago
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