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Salsk061 [2.6K]
2 years ago
10

You have now identified whether crabs are able to see plankton they eat near the ocean floor. Based on your answer, explain how

this information can be used to help the Australian Institute of Marine Biology further to understand the environment of the ocean floor (what other organisms live at the bottom? How are these able to see their food/ attract food to survive?
Chemistry
1 answer:
Murrr4er [49]2 years ago
7 0

They are able to find out more organisms which are present at the bottom of the sea due to their accomplishment.

<h3>How this information can be used to help to understand the environment of the ocean floor?</h3>

This information can be used to help the Australian Institute of Marine Biology further to understand the environment of the ocean floor because they are able to see the seas floor and can find out other creatures as well.

So we can conclude that they are able to find out more organisms which are present at the bottom of the sea due to their accomplishment.

Learn more about crabs here: brainly.com/question/1120881

#SPJ1

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How many milliliter of a solution of 4.00KI are needed to prepare 250.0mL of 0.760 KI
Alexeev081 [22]
Answer:

47.5 mL

Solving:

M1 = 4.00 M

V1 = ?

M2 = 0.760 M

V2 = 0.250 L

---

M1 * V1 = M2 * V2

V1 = ( M2 * V2 ) / M1

V1 = ( 0.760 * 0.250 ) / 4.00

V1 = ( 0.190 ) / 4.00

V1 = 0.0475 L
3 0
4 years ago
Can this reaction take place/does it exist - . CaCl2+CO2+H20----&gt; CaCO3 + 2HCl
mash [69]
Yes the reaction given above does exist
First of all CaCl2 will react with  water to form CaO  and HCl  then it will react with CO2 to form <span>CaCO3 
</span>CaCO3 + 2HCl <span>>>></span>     CaCl2+CO2+H20
so i conclude it does exist
hope it helps
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4 years ago
Please help with Chemistry! Very urgent! I’ll give you 40 points
kvv77 [185]

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8 0
3 years ago
Read 2 more answers
1 C3H8 + 5 O2 --&gt; 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many
UNO [17]

Answer:

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen.  We also know that number of mole is proportional to pressure and volume.  Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law).  So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2.  Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol.  Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent.  Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water.  Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation:

7 0
3 years ago
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