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tino4ka555 [31]
2 years ago
14

Good evening! Does anyone know this? Earth Science

Chemistry
1 answer:
riadik2000 [5.3K]2 years ago
6 0

The fraction of Earth's radius (6371 km) relative to the thickness of the oceanic (7.5 km) and continental crust (35 km) is 0.12 and 0.55, respectively.    

What we know:

  • The average radius of Earth (E) = 6371 km
  • The average thickness of oceanic crust (O) = 7.5 km
  • The average thickness of continental crust (C) = 35 km

We need to convert all the above units from kilometers to miles:

E = \frac{0.6214 mi}{1 km}*6371 km = 3958.9 mi

O = \frac{0.6214 mi}{1 km}*7.5 km = 4.7 mi

C = \frac{0.6214 mi}{1 km}*35 km = 21.7 mi

Now, we can calculate the fraction of Earth's radius relative to each type of crust, with the given equation:

X = \frac{avg. \: thickness}{avg. \: radius} \times 100

  • <u>For the oceanic crust (O)</u>:

X = \frac{4.7 mi}{3958.9 mi}\times 100 = 0.12

  • <u>For the continental crust (C)</u>:

X = \frac{21.7 mi}{3958.9 mi}\times 100 = 0.55

Therefore, the fraction of Earth's radius relative to the oceanic and continental crust is 0.12 and 0.55, respectively.

You can see another example of calculation of fractions of Earth's radius here: brainly.com/question/4675868?referrer=searchResults

I hope it helps you!

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Based on the experiment, it is possible that the solution Z was a buffer and Y another kind of solution. For this reson, pH of the solution Y changes much more than the pH of solution Z changes despite the amount of NaOH added is the same in both solutions.

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How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold? The
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Answer:

The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³

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The percentage by mass of gold in the ore = 0.22%

The density of the ore = 8.0 g/cm³

The price of the gold = $818 per troy ounce

14.6 troy oz = 1.0 pound

1 lb = 454 g

Given that one troy ounce = $818

$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce

1 troy oz = 1.0/14.6 lb

100/818 troy oz =  100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb

1 lb = 454 g

250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g

$100 = 3.8015 g worth of gold

The mass, M, of the ore containing 3.8015 g of gold is given as follows;

0.22% of M = 3.8015 g

0.22/100 × M = 3.8015 g

M = 3.8015 g × 100/0.22 = 1727.933 g

The volume, V, of the ore containing 3.8015 g of gold is given as follows;

Density of ore = Mass of ore/(Volume of ore)

Volume of ore = Mass of ore /(Density of ore)

The density of the ore = 8.0 g/cm³

Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³

Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.

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How many grams of \text{NaCl}NaClstart text, N, a, C, l, end text will be produced from 18.0 \text{ g}18.0 g18, point, 0, start
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Answer:

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0.3239 mole of Cl2 will react with 0.3239 mole of Na to yield 0.3239 mole of NaCl.

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