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ziro4ka [17]
3 years ago
9

3. How many grams are in 9.015 x 1035 atoms of Cobalt?​

Chemistry
1 answer:
pentagon [3]3 years ago
3 0

Answer:

8.822 × 10¹³ g Co

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

9.015 × 10³⁵ atoms Co

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Co - 58.93 g/mol

<u>Step 3: Convert</u>

<u />9.015 \cdot 10^{35} \ atoms \ Co(\frac{1 \ mol \ Co}{6.022 \cdot 10^{23} \ atoms \ Co} )(\frac{58.93 \ g \ Co}{1 \ mol \ Co} ) = 8.82189 × 10¹³ g Co

<u>Step 4: Check</u>

<em>We are given 4 sig figs. Follow sig fig rules and round.</em>

8.82189 × 10¹³ g Co ≈ 8.822 × 10¹³ g Co

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Rank the members of each set of compounds according to the ionic character of their bonds. Most ionic bonds?a) PCl3 PBr3 PF3 Mos
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Explanation:

More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.

(a)   Electronegativity value of P = 2.19

Electronegativity value of Cl = 3.16

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97

Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77

Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79

Since, a P-F bond has the highest electronegativity difference. Therefore, PF_{3} is the most ionic compound and PBr_{3} is the least ionic compound.

(b)   Electronegativity value of B = 2.04

Electronegativity value of N = 3.04

Electronegativity value of C = 2.55

Electronegativity value of F = 3.98

Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94

Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94

Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43

Since, a B-F bond has the highest electronegativity difference. Therefore, BF_{3} is the most ionic compound and NF_{3} is the least ionic compound.

(c)   Electronegativity value of Se = 2.55

Electronegativity value of Te = 2.1

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43

Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88

Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02

Since, a Te-F bond has the highest electronegativity difference. Therefore, TeF_{4} is the most ionic compound and BrF_{3} is the least ionic compound.

6 0
2 years ago
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