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krok68 [10]
3 years ago
13

Which of the following has the smallest radius? Br- I Br I-

Chemistry
2 answers:
Arisa [49]3 years ago
5 0

Answer:

i would say the the first 1

Explanation:

kirill115 [55]3 years ago
4 0

Answer:

Br

Explanation:

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Please help make sure its correct thanks
Wewaii [24]

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g

Then mol iodine (MW=126.9045 g/mol) :

\tt \dfrac{84.1}{126.9045}=0.663

mol ratio of Cobalt and Iodine in the compound :

\tt 0.221\div 0.663=1\div 3

5 0
3 years ago
The nuclear binding energy of one nitrogen-14 atom is (BLANK) x 10A J. Round to 3
d1i1m1o1n [39]

mass defect = mass of constituents - mass of atom

N has 7p and 9n

proton mass ~ 1.00728 amu

neutron mass ~ 1.00866 amu

electron mass ~ 0.000549 amu

Nitrogen mass ~ 14.003074 amu

mass defect = (7*1.00728)-(7*1.00866)-(7*0.000549) - 14.003074

= 0.11235amu

convert to energy, the binding energy = 1.68x10^-11 J


5 0
3 years ago
How many atoms of cobalt are in 4 moles of cobalt?
Anna [14]
<h3>Answer:</h3>

2 × 10²⁴ atoms Co

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

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

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

4 mol Co (Cobalt)

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                               \displaystyle 4 \ mol \ Co(\frac{6.022 \cdot 10^{23} \ atoms \ Co}{1 \ mol \ Co})
  2. Multiply/Divide:                 \displaystyle 2.4088 \cdot 10^{24} \ atoms \ Co

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

2.4088 × 10²⁴ atoms Co ≈ 2 × 10²⁴ atoms Co

4 0
2 years ago
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