A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3

Question

dedylja [7]

4 years ago

8

A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3

g/cm3."

Chemistry

2 answers:

Minchanka [31] · Answer 1 · 2021-12-11T07:47:37+03:00

The mass of gold bar with a volume of $7.379\times{10^{-4}}\;{{\text{m}}^3}$ and density of $19.3\;{\text{g/c}}{{\text{m}}^3}$ is $\boxed{{\text{14241}}{\text{.47 g}}}$

Further explanation:

The property is a unique feature of the substance that differentiates it from the other substances. It is classified into two types:

1. Intensive properties:

These are the properties that depend on the nature of the substance. These don't depend on the size of the system. Their values remain unaltered even if the system is further divided into a number of subsystems. Temperature, refractive index, concentration, pressure, and density are some of the examples of intensive properties.

2. Extensive properties:

These are the properties that depend on the amount of the substance. These are additive in nature when a single system is divided into many subsystems. Mass, energy, size, weight, and length are some of the examples of extensive properties.

Density is considered as the characteristic property of the substance. It is an intensive property. It is defined as the mass per unit volume. It is generally represented by $\rho$ .

The formula to calculate the density of gold is,

${\text{Density of gold}}\left{\rho}}{\text{ = }}\frac{{{\text{Mass of gold}}\left( {\text{M}} \right)}}{{{\text{Volume of gold}}\left({\text{V}}\right)}}$ …… (1)

Rearrange equation (1) to calculate the mass of gold.

${\text{Mass of gold}}=\left({{\text{Density of gold}}}\right)\left({{\text{Volume of gold}}}\right)$ …… (2)

The density of gold is $19.3\;{\text{g/c}}{{\text{m}}^3}$ .

The volume of gold is $7.379\times{10^{-4}}\;{{\text{m}}^3}$ .

The volume is to be converted from ${{\text{m}}^3}$ to ${\text{c}}{{\text{m}}^3}$ . The conversion factor for this is,

${\text{1 }}{{\text{m}}^3}={10^6}{\text{ c}}{{\text{m}}^3}$

So the volume of gold is calculated as follows:

$\begin{gathered}{\text{Volume of gold}}=\left({7.379\times{{10}^{-4}}\;{{\text{m}}^3}}\right)\left({\frac{{{\text{1}}{{\text{0}}^{\text{6}}}{\text{ c}}{{\text{m}}^{\text{3}}}}}{{{\text{ 1}}{{\text{m}}^{\text{3}}}}}}\right)\\=7.379\times{10^2}\;{\text{c}}{{\text{m}}^3}\\\end{gathered}$

The density of gold is $19.3\;{\text{g/c}}{{\text{m}}^3}$ .

The volume of gold is $7.379\times{10^2}\;{\text{c}}{{\text{m}}^3}$ .

Substitute these values in equation (2).

$\begin{gathered}{\text{Mass of gold}}=\left({\frac{{19.3\;{\text{g}}}}{{{\text{1 c}}{{\text{m}}^3}}}}\right)\left({7.379\times{{10}^2}\;{\text{c}}{{\text{m}}^3}}\right)\\={\mathbf{14241}}{\mathbf{.47 g}}\\\end{gathered}$

Learn more:

1. Calculation of volume of gas: brainly.com/question/3636135

2. Determine how many moles of water produce: brainly.com/question/1405182

Answer details:

Grade: Middle School

Subject: Chemistry

Chapter: Density

Keywords: density, mass, volume, density of gold, mass of gold, volume of gold, 14241.47 g, 19.3 g/cm3, conversion factor, mass per unit volume, characteristic property, intensive, extensive, same, additive.

IRINA_888 [86] · Answer 2 · 2021-12-11T05:42:56+03:00

<h3><em>Further Explanation</em></h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

$\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}$

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

<em>Known variable</em>

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

<em>Asked</em>

the mass of a gold bar

<em>Answer</em>

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram