The anode is the electrode where the oxidation occurs.

Cathode is the electrode where the reducction occurs.

Equations:

Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V

The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.

Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72

Answer: 2.72 V

3 0

3 years ago

Which of the following objects would require the use of a large-scale model A. An atom B. A car engine C. A frog's heart D. The

levacccp [35]

Answer:

A. An atom

Explanation:

A large scale model is one that is bigger than the object itself. These are needed to visualise structures and behaviours that we cannot normally see. A large scale model is needed for small objects, such as an atom. Larger objects do not need<em> even bigger</em> models

4 0

3 years ago

How many atoms are in 1.4 moles of H3PO4?

sdas [7]

<h3>Answer:</h3>

8.4 × 10²³ atoms H₃PO₄

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Moles
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>

[Given] 1.4 moles H₃PO₄

[Solve] atoms H₃PO₄

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.4 \ mol \ H_3PO_4(\frac{6.022 \cdot 10^{23} \ atoms \ H_3PO_4}{1 \ mol \ H_3PO_4})$
[DA] Multiply [Cancel out units]: $\displaystyle 8.4308 \cdot 10^{23} \ atoms \ H_3PO_4$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

8.4308 × 10²³ atoms H₃PO₄ ≈ 8.4 × 10²³ atoms H₃PO₄

3 0

3 years ago