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Rashid [163]
2 years ago
9

2. The density of helium is 1.78 X 104 g/cm. What is this density in Dg/um??

Chemistry
1 answer:
Zepler [3.9K]2 years ago
4 0

Answer:

d=1.78\times 10^{-7}\ Dg/\mu m^3

Explanation:

Given,

The density of Helium is 1.78\times 10^4\ g/cm^3

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3

So, the density of Helium is equal to 1.78\times 10^{-7}\ Dg/\mu m^3.

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stiv31 [10]

Answer:

4.13×10²⁷ molecules of N₂ are in the room

Explanation:

ideal gases Law → P . V = n . R . T

Pressure . volume = moles . Ideal Gases Constant . T° K

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Let's determine the volume of the room:

18 ft . 18 ft . 18ft = 5832 ft³

We convert the ft³ to L → 5832 ft³ . 28.3L / 1 ft³ = 165045.6 L

1 atm .  165045.6 L = n . 0.082 L.atm/mol.K . 293K

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6869.4 mol . 6.02×10²³ = 4.13×10²⁷ molecules

4 0
2 years ago
An unknown compound is processed using elemental analysis and found to contain 117.4g of platinum 28.91 carbon and 33.71g nitrog
dlinn [17]

Answer:

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Pt = 117.4 / 195 = 0.602

C = 28.91 / 12 = 2.409

N = 33.71 / 14 = 2.408

Divide by the smallest

Pt = 0.602 / 0.602 = 1

C = 2.409 / 0.602 = 4

N = 2.408 / 0.602 = 4

The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄

From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.

8 0
2 years ago
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Answer:

can you show us the list

Explanation:

8 0
2 years ago
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18 moles of water are produced in the above reaction.
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3 0
2 years ago
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Each A atom is adjacent to 3 B atoms. What is the A-C-B bond angle?
Nutka1998 [239]
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3 0
3 years ago
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