Calculate the oxidation number of the iodine (I) in each compound: HIO4 = I2 = NaI = HIO3 =

2 answers:

8 0

1) in periodic acid (HIO₄), iodine has oxidation number +7, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 4 · (-2) = 0.
x = +7.

2) in molecule of iodine (I₂), iodine has oxidation number 0, because iodine is nonpolar molecule.

3) in sodium iodide (NaI), iodine has oxidation number -1, sodium has oxidation number +1:
+1 + x = 0.
x = -1.

4) in iodic acid (HIO₃), iodine has oxidation number +5, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 3 · (-2) = 0.
x = +5.

Tems11 [23]2 years ago

6 0

Answer:

the answer has the plus sign in it...

-1

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A sample of CO2 weighing 86.34g contains how many molecules?

irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

Chemistry - Atomic Structure

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

Explanation:

Step 1: Define

86.34 g CO₂

Step 2: Identify Conversion

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Convert

$86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} )$ = 1.18141 × 10²⁴ molecules CO₂