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Snowcat [4.5K]
3 years ago
12

]If a sample of pure iodine contains 4.69 x 1022 atoms of iodine, how many moles of iodine are in the sample? A=0.08 B=0.13 C=7.

78 D=12.84
Chemistry
2 answers:
Lorico [155]3 years ago
6 0
It is B, and also for a moment I didn't understand that 4.69 x 10^22. I almost did this whole problem wrong.
Dmitrij [34]3 years ago
5 0
The correct answer should be B
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51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
What would be the final volume of the new solution if the 0.2 m solution on the left were diluted to 0.04 m? ml quizley?
marin [14]
Missing question: volume of <span>solution on the left is 10 mL.
V</span>₁(solution) = 10 Ml.
c₁(solution) = 0.2 M.<span>
V</span>₂(solution) = ?.<span>
c</span>₂(solution) = 0.04 M.<span>
c</span>₁ - original concentration of the solution, before it gets diluted.<span>
c</span>₂ - final concentration of the solution, after dilution.<span>
V</span>₁ - <span>volume to be diluted.
V</span>₂ - <span>final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂<span>.
</span>10 mL · 0.2 M = 0.04 M · V₂.
V₂(solution) = 10 mL · 0.2 M  ÷ 0.04 M.
V₂(solution) = 50 mL.<span>

</span>
3 0
3 years ago
What’s 73m equal to in dm
MaRussiya [10]

Answer:730

Explanation:

6 0
3 years ago
Please help, and answer as much as you can. ​
n200080 [17]

Answer:

∴ Fractional distillation is the technique used to separate the fraction

Explanation:

A: Refinery gas

B: Gasoline fraction

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5 0
2 years ago
When a mixture containing cations of Analytical Groups I–III is treated with H2S in acidic solution, which cations are expected
Fantom [35]

Answer:

Analytical Groups I and II

Explanation:

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Whether or not a type of reaction like that takes place can be determined by utilizing the solubility principles for common ionic solids.

8 0
2 years ago
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