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adelina 88
6 days ago
5

Solving applied density problems Mr. Auric Goldfinger, criminal mastermind, intends to smuggle several tons of gold across Inter

national borders by disguising it as lumps of iron ore. He commands his engineer minions to form the gold into little spheres with a diameter of exactly 6 cm and paint them black However, his chief engineer points out that customs officials will surely notice the unusual weight of the "iron ore" if the balls are made of solid gold (density 19.3 g/cm'). He suggests forming the gold into hollow balls Instead (see sketch at right), so that the fake "Iron ore" has the same density as real iron ore (5.15 g/cm'). One of the balls of fake iron ore," sliced in half Calculate the required thickness of the walls of each hollow lump of iron ore. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits. OP 0.0 X 5
Chemistry
1 answer:
KiRa [971]6 days ago
8 0

Answer:

thickness is 0.29 cm

Explanation:

To create a fake iron ball out of gold, we must ensure that its mass matches that of the iron ball. Therefore, we first find the volume of the iron ball using the provided diameter, applying the formula of 4/3 pi r^3.

Given the diameter d = 6 cm; thus, the radius r = 3 cm (d/2).

We calculate the volume of the iron ball: 4/3 * 3.14 * 3^3 = 113.04 cm^3.

The corresponding mass of the iron ball is the volume multiplied by its density: 113.04 * 5.15 g/cm^3 = 582.156 g.

This value represents the mass for the gold ball; now we determine the volume of the gold ball using its density.

Volume of gold ball = mass of gold ball/density of gold = 582.156 g/19.3 g/cm^3 = 30.1635 cm^3.

So this volume must correspond to a hollow sphere with an outer radius R = 3 cm and an unknown inner radius r.

Volume of the hollow ball can be represented as: 4/3 pi [R^3 - r^3].

Thus, 30.1635 cm^3 = 4/3 pi [3^3 - r^3].

30.1635 * 3/(4 * 3.14) = 27 - r^3.

Simplifying gives 7.2046 = 27 - r^3, resulting in r^3 = 19.7954.

Therefore, r = 2.7051 cm.

This indicates the thickness is the outer radius minus the inner radius: 3 - 2.7051 = 0.2949 cm.

Rounding to two significant figures yields

the thickness = 0.29 cm.

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A certain alcoholic beverage contains only ethanol (C2H6O) and water. When a sample of this beverage undergoes combustion, the e
castortr0y [923]

Response:

9.606 g

Clarification:

Step 1: Write the balanced combustion equation

C₂H₆O(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(g)

Step 2: Determine the moles for 11.27 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

11.27 g × (1 mol/18.02 g) = 0.6254 mol

Step 3: Find the moles of C₂H₆O that produced 0.6254 moles of H₂O

The ratio of C₂H₆O to H₂O is 1:3. Thus, the moles of C₂H₆O are 1/3 × 0.6254 mol = 0.2085 mol

Step 4: Calculate the mass for 0.2085 moles of C₂H₆O

The molar mass of C₂H₆O is 46.07 g/mol.

0.2085 mol × 46.07 g/mol = 9.606 g

7 0
10 days ago
A sample of neon gas at STP is allowed to expand into an evacuated vessel. What is the sign of work for this process?
Alekssandra [968]

Answer:

The work done in this process will be considered Negative.

Explanation:

The energy transferred by the system to the environment is negative

Therefore, if work is done on the system, it is labeled as positive. Conversely, when work is done by the system, it is regarded as negative.

In this scenario, the argon gas is expanding, and the work is exerted by the system into the surroundings (container), making the sign Negative.

Thus, the result for the work pertaining to this process will carry a Negative sign.

3 0
9 days ago
What volume of 0.550 M KBr solution can you make from 100.0 mL of 2.50 M KBr?
castortr0y [923]
M1V1 = M2V2
(2.50)(100.0) = (0.550)V2
V2 = 455mL

From 100.0 mL of 2.50 M KBr, you can prepare 455 mL of 0.550 M solution.
5 0
14 days ago
Read 2 more answers
A compound composed of only carbon and chlorine is 85.5% chlorine by mass. propose a lewis structure for the lightest of the pos
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The visual representation is displayed in the following image.

For calculations, consider 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; this represents the mass of chlorine.

m(C) = 100 g - 85.5 g.

m(C) = 14.5 g; indicating the mass of carbon.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; this is the quantity of chlorine.

n(C) = 14.5 g ÷ 12 g/mol.

n(C) = 1.21 mol; this is the quantity of carbon.

n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.

The compound in question is identified as dichlorocarbene CCl₂.

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Acetone major species present when dissolved in water
Alekssandra [968]

The compound is acetone ( CH₃-CO-CH₃)


Explanation:


1) Acetone is represented as CH₃-CO-CH₃.


2) This is a molecule formed by covalent bonds.


3) When it dissolves, compounds with covalent bonds remain as individual molecules, indicating that the primary species in the solution are the molecules themselves, which are surrounded (solvated) by water molecules.


In contrast, ionic compounds ionize. For example, when NaCl dissolves in water, it completely breaks down into ions, hence the predominant species are the ions Na⁺ and Cl⁻, rather than the NaCl formula.


This leads to the conclusion that: when acetone dissolves in water, the primary components are the acetone molecules (there is no need to mention that water molecules are in the solution, as that isn't the question's focus).



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