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Elena L [17]
3 years ago
13

Describe the oceanic crust NO LINKS

Chemistry
2 answers:
klasskru [66]3 years ago
7 0

Answer:

Oceanic crust, the outermost layer of Earth’s lithosphere that is found under the oceans. Oceanic crust is constantly formed at mid-ocean ridges, where tectonic plates are tearing apart from each other.

Explanation:

torisob [31]3 years ago
4 0

Answer:

oceanic crust is the relatively thin part of the earth's crust which underlies the ocean basins. It is geologically young compared with the continental crust and consists of basaltic rock overlain by sediments

Explanation:

I hope this helped ;)

You might be interested in
Explain why one prevents erosion and the other makes erosion worse.​
Studentka2010 [4]
Eroded sediments from further upstream clogged Jakarta’s rivers and canals, causing them to overflow. Similar erosion-related floods have occurred in many other countries, such as Colombia, India, the Philippines and Democratic Republic of the Congo.
6 0
3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0
3 years ago
How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --&gt; __ AlCl3 + __
kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0
2 years ago
Write a reflection about your learning in this unit. Your reflection should be at least 3 sentences. Use the following sentence
Vera_Pavlovna [14]

Answer:

This unit has encouraged a deeper understanding of the world and it's guiding principles. While it was initially challenging for me to determine if a change was physical or chemical, this unit provided me with the information necessary to determine the type. With this knowledge, I can now interrelate with other properties and believe that this new ability will assist in future units as well. Thank you!

Explanation:

5 0
2 years ago
3. How many moles of silver is 8.46x1024 atoms of<br> silver?
ioda

Answer:

<h2>14.05 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{8.46 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 14.053156

We have the final answer as

<h3>14.05 moles</h3>

Hope this helps you

6 0
2 years ago
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