Question

A swimming pool measures 45.0 meters by 30.0 meters. How many grams of water are needed to fill the pool, whose average depth is 7.2 feet? Assume the density of water to be 1.0 g/mL. Report answer to 3 significant figures. Do not enter “grams” as part of your answer. Use “E” for scientific notation.

Answer 1

Answer:
mass of water = 2962656000 = 2.963 E9 grams

Explanation:
1 ft = 0.3048 m
Therefore:
7.2 ft = 2.19456 m

First, we will get the volume of the swimming pool as follows:
The swimming pool has a rectangular shape, this means that:
volume of pool = length * width * depth
volume of pool = 45 * 30 * 2.19456 = 2962.656 m^3

Now, we will get the mass of needed water as follows:
density = mass / volume
we have:
density = 1 g/ml
volume = 2962.656 m^3 = 2962656000 ml
Substitute to get the mass of water as follows:
mass of water = density * volume = 1 * 2962656000
mass of water = 2962656000 = 2.963 * 10^9 grams

Hope this helps :)

Answer 2

The mass of water to fill the swimming pool of length 45 m, width 30 m and a depth of 72 ft is $\boxed{2.963{\text{E}}9}$

Further explanation:

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

The formula to calculate the density of water is,

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

The swimming pool is in the shape of a cuboid. Its length, width, and depth are 45 m, 30 m and 7.2 ft.

Firstly, we have to convert the depth of the pool from feet to meters. The conversion factor for this is,

 ${\text{1 ft}}={\text{0}}{\text{.3048m}}$

So the depth of the pool is calculated as follows:

$\begin{gathered}{\text{Depth of pool}}=\left({7.2\;{\text{ft}}}\right)\left({\frac{{{\text{0}}{\text{.3048m}}}}{{{\text{1 ft}}}}}\right)\\={\text{2}}{\text{.19456m}}\\\end{gathered}$

The formula to calculate the volume of the pool is as follows:

${\text{Volume of pool}}=\left({{\text{length of pool}}}\right)\left({{\text{width of pool}}} \right)\left( {{\text{depth of pool}}}\right)$          …… (2)

The length of the pool is 45 m.

The width of the pool is 30 m.

The depth of the pool is 2.19456 m.

Substitute these values in equation (2).

$\begin{gathered}{\text{Volume of pool}}=\left({{\text{45m}}}\right)\left({{\text{30m}}}\right)\left({{\text{2}}{\text{.19456m}}}\right)\\=2962.656\;{{\text{m}}^{\text{3}}}\\\end{gathered}$

As the pool is filled with water so the volume of water is the same as that of the swimming pool and therefore the volume of water is $2962.656\;{{\text{m}}^{\text{3}}}$.

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

${\text{Mass of water}}=\left({{\text{Density of water}}}\right)\left({{\text{Volume of water}}}\right)$                 ……. (3)

The conversion factor to convert the volume of water in mL is as follows:

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

So the volume of water is calculated as follows:

$\begin{gathered}{\text{Volume of water}}=\left({2962.656\;{{\text{m}}^{\text{3}}}}\right)\left({\frac{{{{10}^6}\;{\text{mL}}}}{{1\;{{\text{m}}^3}}}}\right)\\=2962656000\;{\text{mL}}\\\end{gathered}$

The density of water is 1 g/mL.

The volume of water is $2962656000\;{\text{mL}}$.

Substitute these values in equation (3).

$\begin{gathered}{\text{Mass of water}}=\left({\frac{{{\text{1 g}}}}{{{\text{1 mL}}}}}\right)\left({{\text{2962656000 mL}}}\right)\\={\text{2962656000g}}\\\end{gathered}$

So ${\text{2962656000 g}}$ water is needed to fill the swimming pool. Its value up to 3 significant figures is $2.963\times{10^9}$ and it is written as $2.963{\text{E}}9$ in scientific notation.

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, swimming pool, length, width, depth, water, volume of water, density of water, significant figures, scientific notation, 2.963E9, 2962656000 g.