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zubka84 [21]
3 years ago
6

That's just the tip of the iceberg" is a popular expression you may have heard. It means that what you can see is only a small p

art of the overall problem. As the diagram shows, most of an iceberg is actually out of sight, below the water level. Based on this diagram, what is the most likely density of the iceberg? (Assume a density of 1.03 g/mL for seawater.)
A. 0.88 g/cc
B. 1.23 g/cc
C. 0.23 g/cc
D. 4.14 g/cc
Chemistry
1 answer:
puteri [66]3 years ago
5 0
Answer:

B 1.23 g/cc

Explanation:
For something to float on seawater, the density must be less than 1.03 g/mL. If the object sinks, the density is greater than 1.03 g/mL.

Let’s examine the answer choices. Keep in mind, the ice berg is mostly below the water level.

A. 0.88 g/cc
This is less than 1.03 g/cc, which would result in floating.

B. 1.23 g/cc
This is the best answer choice. The iceberg is mostly beneath the water, but some of it is exposed. The density is greater than 1.03 g/mL, but not so much greater that it would immediately sink.

C. 0.23 g/cc
This is less than 1.03 g/cc, which would produce floating.

D. 4.14 g/cc
This is much greater than 1.03 g/cc and the result would be sinking.
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zhuklara [117]

To solve this problem, we should recall that the change in enthalpy is calculated by subtracting the total enthalpy of the reactants from the total enthalpy of the products:

ΔH = Total H of products – Total H of reactants

You did not insert the table in this problem, therefore I will find other sources to find for the enthalpies of each compound.

ΔHf CO2 (g) = -393.5 kJ/mol

ΔHf CO (g) = -110.5 kJ/mol

ΔHf Fe2O3 (s) = -822.1 kJ/mol

ΔHf Fe(s) = 0.0 kJ/mol

Since the given enthalpies are still in kJ/mol, we have to multiply that with the number of moles in the formula. Therefore solving for ΔH:

ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0 kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>

ΔH = <span>795.2 kJ</span>

3 0
3 years ago
How many minutes will it take to plate out 4.50 g of Cu from a solution of Cu(NO3)2 (aq) onto the cathode of an electrolytic cel
ipn [44]

Answer:

It will take 28.5 minutes

Explanation:

<u>Step 1: </u>Data given

Mass of Cu = 4.50 grams

8.00 A of current are used

Molar mass of Cu = 63.5 g/mol

Step 2: Calculate time needed

Cu2+ →Electricity → Cu

we notice a flow of 2 electrons ⇒ This means the Faraday constant = 2F

Since Molar mass of Cu is 63.5 g/mol

63.5 grams of Cu is deposited by 2*96500 C

4.50 grams of Cu ((2*96500)/63.5)  * 4.50 = 13677.17 C

Q = It

13677.17 = 8t*60 seconds

t = 28.5 minutes

3 0
3 years ago
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babymother [125]

Answer:

Option B.

Explanation:

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2C₂H₆  +  7O₂   →   4CO₂  +  6H₂O

So 2 moles of ethane react with 7 moles of oxygen to make 4 moles of dioxide and 6 moles of water.

Then 2 moles of ethane will produce 4 moles of CO₂

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Answer:

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Explanation:

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