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12345 [234]
3 years ago
6

What is the answer ?

Chemistry
2 answers:
Diano4ka-milaya [45]3 years ago
8 0
Aaa no one has a chance of a meeting today or Friday at
insens350 [35]3 years ago
6 0

Answer: Its not popping up for me *-*?

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Sveta_85 [38]

Answer:

Brittle

Explanation:

because thas for non metals

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3 years ago
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At STP conditions, 11 g of SO, (64.06 g/mole) have a volume of
raketka [301]

Answer:

3.8 L

Explanation: Message me for explanation.

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The scattering of light by a colloid is called the Brownian lighting, Tyndall effect, colloidal scattering, or aggregate reflect
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The correct answer is the Tyndall effect. This is also known as the Tyndall scattering. It is the light scattering by the particles in a colloid or in a suspension. This phenomenon is used to determine size and density of particles in colloidal matter.
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3 years ago
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What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?
ioda

N₂O is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen.

Empirical formula can be calculated by

Suppose we have 100 g of the substance. That indicates that it has 36.69 grams of oxygen and 63.61 grams of nitrogen.

Masses transformed into moles:

Formula used

Given mass/ Molar mass

14.01 g contains 1 mol of N

So 63.61 g of N contains moles is equals to

(1 mol N / 14.01 g N) 63.61 g N = 4.540 mol N

Similarly

16 g of O contains 1 mole of O

36.69 g of O contains moles is equals to

(1 mol O / 16.00 g O) 36.69 g O = 2.293 mol O

Divide by the smallest to normalize:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

Therefore, there are roughly twice as many N as O atoms. N2O is the empirical formula as a result.

Ratio is basically 2:1

Hence, N₂O is the empirical formula of an oxide of nitrogen

Learn more about Empirical Formula here brainly.com/question/27873410

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2 years ago
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It’s static friction
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