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Naily [24]
3 years ago
12

Prospectors are considering searching for gold on a plot of land that contains 2.45 g of gold per bucket of soil. If the volume

of the bucket is 5.25L, how many grams of gold are there likely to be in 1.0 cubic feet of soil
Chemistry
1 answer:
makvit [3.9K]3 years ago
8 0

Answer:

13.2 g of gold

Explanation:

We'll begin by converting 5.25 L to ft³.

This can be obtained as follow:

Recall:

1 L = 0.0353 ft³

Therefore,

5.25 L = 5.25 × 0.0353

5.25 L = 1.85×10¯¹ ft³

From the question given above,

2.45 g of gold is present in 5.25 L ( i.e 1.85×10¯¹ ft³) of soil.

Therefore, Xg of gold will be present in 1 ft³ of soil i.e

Xg of gold = 2.45/1.85×10¯¹

Xg of gold = 13.2 g

Therefore, 13.2 g of gold is present in 1 ft³ of the soil.

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HELP IM AB TO CRY
Lemur [1.5K]

Answer:

1.23 × 10³ N

Explanation:

Step 1: Given and required data

  • Mass of the person (m): 125 kg
  • Acceleration due to the gravitational force (g): 9.81 m/s²

Step 2: Calculate the force acting between the Earth and a 125-kg person standing on the surface of the Earth

We will use Newton's second law of motion.

F = m × g

F = 125 kg × 9.81 m/s²

F = 1.23 × 10³ N

8 0
3 years ago
The force of attraction between a divalent cation and a divalent anion is 1.64 x 10-8 N. If the ionic radius of the cation is 0.
Elden [556K]

The radius of the anion is 7.413 nm

<h3>How to calculate the force of attraction between charges</h3>

The force of attraction (F) is given by the formula:

  • F = (1/4π∈r²)(Zc*e)(Za*e)

where:

∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m

Zc = charge on the cation = +2

Zc = charge on the anion = -2

e = charge on an electron = 1.602 * 10⁻¹⁹ C

r = interionic distance

r = rc + ra

where rc and ra are the radius of the cation and anion respectively

F = 1.64 * 10⁻⁸ N

Therefore based on the equation of force of attraction:

1.64 *10⁻⁸ = [1/4π(8.85*10⁻¹⁵)r²](2 * 1.602*10⁻¹⁹)²

r² = 5.63 * 10⁻¹⁷

r = 7.50 nm

Since r = rc + ra

where rc = 0.087 nm

thus, ra = r - rc = 7.50 - 0.087

ra = 7.413 nm

Therefore, the radius of the anion is 7.413 nm

Learn more about ionic radius at: brainly.com/question/2279609

6 0
2 years ago
Guys please answer this
iren [92.7K]

Answer:

Mass = 96 g

Explanation:

Given data:

Number of moles of C = 8 mol

Mass of C in gram = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of C = 12 g/mol

8 mol = mass / 12 g/mol

Mass = 8 mol × 12 g/mol

Mass = 96 g

6 0
3 years ago
What happens to a liquid when you keep on cooling it until it changes?
Gnesinka [82]

Answer:

it turns into a solid

6 0
3 years ago
4. ¿Cuál es la resistencia de un alambre de
Artemon [7]

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:

R=

El resultado es:

R=0.0438 Ω

Explanation:

5 0
3 years ago
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