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nikitadnepr [17]
3 years ago
10

An estimate of the amount of water frozen at the Moon's south pole is 100,000 m3. If this deposit was spread over an area measur

ing 160 m by 125 m, how many meters deep would the deposit be?
Chemistry
1 answer:
jolli1 [7]3 years ago
5 0

5m

Explanation:

Given parameters:

Volume of frozen water = 100000m³

Area of the deposit = 160m by 125m

Unknown;

depth of the deposit = ?

Solution:

Volume is the amount of space occupied by a body.

The volume of this ice is given by;

      Volume of ice = length x breadth x depth

Since the unknown is depth, we simply solve for it;

          Depth = \frac{100000}{160 x 125} =  5m

The ice is about 5m deep

learn more:

Density and volume brainly.com/question/8441651

#learnwithBrainly

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Provide a specific example the relationship between processing, structure, and properties of an engineering material. Be specifi
k0ka [10]

Answer:

An example of engineering material, <em><u>are plastics,</u></em> they are derived from organic, natural materials, such as cellulose, coal, natural gas, salt and, of course, oil. Oil is a complex mixture of thousands of compounds and must be processed before being used.

Explanation:

Plastic production begins with distillation at a refinery, where crude oil is separated into groups of lighter components, called fractions. Each fraction is a mixture of hydrocarbon chains (chemical compounds formed by carbon and hydrogen) that differ in terms of the size and structure of their molecules. One of those fractions, naphtha, is the essential compound for the production of plastic.

Two main processes are used to make plastic: polymerization and polycondensation, and both require specific catalysts. In a polymerization reactor, monomers like ethylene and propylene join to form long polymer chains. Each polymer has its own properties, structure and dimensions depending on the type of basic monomer that has been used.

5 0
3 years ago
What causes the shielding effect to remain constant across a period?
Ivanshal [37]
Electrons are added to the same principal energy level.
5 0
3 years ago
If 120.3 mL of water is shaken with oxygen gas at 2.1 atm, it will dissolve 0.0043 g O2. Estimate the Henry's law constant for t
nikklg [1K]

<u>Answer:</u> The Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{O_2}=K_H\times p_{O_2}

where,

K_H = Henry's constant = ?

C_{O_2} = solubility of oxygen gas = 0.0043g/120.3mL

p_{O_2 = partial pressure of oxygen gas = 2.1 atm

Putting values in above equation, we get:

0.0043g/120.3mL=K_H\times 2.1atm\\\\K_H=\frac{0.0043g}{120.3mL\times 2.1atm}=1.702\times 10^{-5}g/mL.atm

Hence, the Henry's law constant for oxygen gas in water is 1.702\times 10^{-5}g/mL.atm

7 0
2 years ago
Which of the following electron configurations are written incorrectly?
Lynna [10]

Answer:

The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.

Explanation:

The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.

There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.

All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".

In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.

The other options are correctly written.

3 0
2 years ago
Given that you started with 28.5 g of K3PO4, how many grams of KNO3 can be<br> produced?
Irina18 [472]

Mass of KNO₃ : = 40.643 g

<h3>Further explanation</h3>

Given

28.5 g of K₃PO₄

Required

Mass of KNO₃

Solution

Reaction(Balanced equation) :

2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃

mol K₃PO₄(MW=212,27 g/mol) :

= mass : MW

= 28.5 : 212,27 g/mol

= 0.134

Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :

= 6/2 x mol K₃PO₄

= 6/2 x 0.134

= 0.402

Mass of KNO₃ :

= mol x MW KNO₃

= 0.402 x 101,1032 g/mol

= 40.643 g

8 0
2 years ago
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