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zubka84 [21]
2 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]2 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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Answer:

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Explanation:

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Which of the following describes the electron sharing between hydrogen and fluorine?
EastWind [94]
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We know both the elements involved in bonding are non-metals and the primary type of bonds involved in non-metals are covalent bonds. Covalent bonds are formed when two atoms share one or more electrons; thus we know that whatever the number of electrons shared, it has to be equal for both. We can eliminate choices A and B.
Next, we understand that it is easier for one atom to be shared among the two, rather than the 7. First, because Hydrogen needs only 1 electron to be stable and would require energy to compensate the remaining 6. Second, electrons are attracted towards the nucleus so it is inefficient to try and share 7 electrons when 1 provides the same amount of stability to Fluorine.
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4 0
2 years ago
Read 3 more answers
A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me
DiKsa [7]

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0
2 years ago
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Part a how many grams of xef6 are required to react with 0.579 l of hydrogen gas at 6.46 atm and 45°c in the reaction shown belo
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First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.

PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂

The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
6 0
2 years ago
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