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Anon25 [30]
3 years ago
12

In a oxidation–reduction reaction, oxidation and reduction must occur simultaneously. Is this statement true or false?

Chemistry
1 answer:
bogdanovich [222]3 years ago
5 0

Answer:

True

Explanation:

Oxidation and reduction always occur simultaneously. Gave example when a substance is oxidized or reduced? 1) The substance gaining oxygen or losing electron is oxidized. 2) The substance losing oxygen ( or gaining electron) is reduced. hope this helps you :)

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The state with the lowest possible energy consistent with the state of maximum disorder

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Hydrogen peroxide slowly decomposes in light:
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Choose the aqueous solution below with the highest boiling point. These are all solutions of nonvolatile solutes and you should
o-na [289]

The question is incomplete, the complete question is;

Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. 0.100 m C6H12O6 0.100 m AlCl3 0.100 m NaCl 0.100 m MgCl2 They all have the same boiling point.

Answer:

AlCl3 0.100 m

Explanation:

Let us remember that the boiling point elevation is given by;

ΔTb = Kb m i

Where;

ΔTb = boiling point elevation

Kb = boiling point constant

m = molality of the solution

i = Van't Hoff factor

We can see from the question that all the solutions possess the same molality, ΔTb now depends on the value of the Van't Hoff factor which in turn depends on the number of particles in solution.

AlCl3 yields four particles in solution, hence ΔTb is highest for AlCl3 . The solution having the highest value of ΔTb also has the highest boiling point.

8 0
2 years ago
How many molecules of co2 in 97.3 grams of co2
s2008m [1.1K]

Answer:

1.33 × 10²⁴ molecules CO₂

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

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

Explanation:

<u>Step 1: Define</u>

97.3 g CO₂

<u>Step 2: Define conversions</u>

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

<u>Step 3: Convert</u>

97.3 \ 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} ) = 1.33138 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules.</em>

1.33138 × 10²⁴ molecules CO₂ ≈ 1.33 × 10²⁴ molecules CO₂

8 0
3 years ago
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