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2 years ago

4. ¿Cuál es la resistencia de un alambre de

Chemistry

1 answer:

Artemon [7]2 years ago

5 0

Answer:

R=0.0438 Ω

Explicación:

1) Hallar el área o sección del conductor de cobre, usando esta fórmula:

A=π.r² (Pi x radio al cuadrado)

Debido a que conocemos el diámetro (1.5mm) su radio es la mitad de esto es decir 0.75mm, y lo sustituimos en la fórmula:

A=π.(0.75mm)²

A=π(0.5625mm²)

A=1.7671mm²

2) La resistividad del cobre es: rho = 0,0172 y la incluimos en la fórmula siguiente:

R=p

R=0,0172Ω x

Simplificamos:

El resultado es:

R=0.0438 Ω

Explanation:

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How many molecules are in 85g of silver nitrate?

maksim [4K]

<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<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>

Reading a Periodic Table
Writing Compounds
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>

85 g AgNO₃ (silver nitrate)

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

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})$
Multiply/Divide: $\displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3$

<u>Step 4: Check</u>

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

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

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