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Triss [41]
3 years ago
15

What is the main difference between Lewis structure and Bohr models?

Chemistry
1 answer:
AlekseyPX3 years ago
5 0

Answer:

Lewis Diagrams. Elemental properties and reactions are determined only by electrons in the outer energy levels. Electrons in completely filled energy levels are ignored when considering properties. Simplified Bohr diagrams which only consider electrons in outer energy levels are called Lewis Symbols.

Explanation:

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What is the mass of 2.90 ×1022 molecules of NaOH (Molar mass = 40.0 g/mol)?
ivann1987 [24]
40.0 g ( 1 mole ) --------------- 6.02x10²³ molecules
      ? ? --------------------------- 2.90x10²² molecules

mass = 2.90x10²² * 40.0 /  6.02x10²³

mass = 1.16x10²⁴ / 6.02x10²³

mass = 1.9269 g

hope this helps!

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A sample liquid is heated in a closed container until it changes to gas.what happens to the size of the particles in the sample
murzikaleks [220]
Nothing happens to these particles when it comes to size however if it were to be speed, the sample would increase.
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Explanation:

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How much space does 1 mole of any gas occupy?
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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
bezimeni [28]

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

Explanation:

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.

5 0
3 years ago
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