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3 years ago

Jack has a rock the rock has a mass of 14 G and the volume of 12 cm power of 2 that what is the Density of the rock?

Chemistry

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

Given parameters:

Mass of rock = 14g

Volume of rock = 12cm³

Unknown:

Density of the rock = ?

Density of a substance is the amount of substance it contains per unit volume. Mathematically, this can be expressed as;

Density = $\frac{Mass}{Volume}$

Now simply input the parameters and solve;

Density = $\frac{14}{12}$ = 1.17g/cm³

Density of the rock that Jack has is 1.17g/cm³

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3 years ago

Use the balanced equation to solve the problem.

harkovskaia [24]

Answer:

4.20 moles NF₃

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2 years ago

What are the uses of solution?

Ilia_Sergeevich [38]

Answer:

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Turpentine as a solvent are used in the production of paints, inks and dyes. ↔Water as a solvent is used in the making of food, textiles, soaps and detergents. ↔Alloys are solid solutions that are used in the manufacture of cars, aerospace and other vehicles.

Explanation:

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3 years ago

What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha

sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

Given:

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

To determine:

The mean free path, λ

Calculation:

The mean free path is related to the collision cross section by the following equation:

$\lambda =\frac{1}{n\sigma }------(1)$

where n = number density

$n = \frac{P}{kT}-----(2)$

Substituting for P, k and T in equation (2) gives:

$n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\ m^{3}$

Next, substituting for n and σ in equation (1) gives:

$\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m$

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3 years ago