Chapter 1, Problem 49QAP

To determine

(a)

The limit of sodium intake per day for adult in unit of kilogram.

Answer to Problem 49QAP

The sodium intake limit for adult is $2.5\times {10}^{-3}\text{kg}$.

Explanation of Solution

Given info:

An adult cannot take more than $\text{2500 mg}$ per day.

Introduction:

The International System of Units is the most widely used system of measurement. It has seven basic unit and twenty-two derived units.

Convert the unit of sodium limit intake from milligram to kilogram.

  $\begin{array}{c}\text{2500 mg}\text{=}\text{2500 mg}\times \frac{1\text{kg}}{{10}^6\text{g}}\\ =2.5\times {10}^{-3}\text{kg}\end{array}$

Conclusion:

The sodium intake limit for adult is $2.5\times {10}^{-3}\text{kg}$.

To determine

(b)

The concentration of cholesterol in a cup of milk in unit of ${\text{kgm}}^{-3}$ and $\text{mg/mL}$.

Answer to Problem 49QAP

The concentration of cholesterol in a cup of milk is $\text{0}{\text{.146 kgm}}^{-3}$ and $\text{0}\text{.146 mg/mL}$.

Explanation of Solution

Given info:

A $\text{240}\text{mL}$ of milk contains $\text{35}\text{mg}$ of cholesterol.

Introduction:

The International System of Units is the most widely used system of measurement. It has seven basic unit and twenty-two derived units.

Formula used:

Formula to find the concentration of liquid is,

  $\text{Concentration =}\frac{\text{Amount}\text{of}\text{substance}}{\text{Volume}\text{of}\text{container}}\text{Equation-1}$

Find the concentration of milk in unit of $\text{mg/mL}$,

  $\begin{array}{c}\text{Concentration =}\frac{\text{35}\text{mg}}{\text{240}\text{mL}}\\ =0.146\text{mg/mL}\end{array}$

Find the concentration of milk in unit of ${\text{kgm}}^{-3}$

First find the amount of milk in unit of kg.

Convert the unit from milligram to kilogram,

  $\begin{array}{c}\text{35}\text{mg}=\text{35}\text{mg×}\frac{\text{1}\text{g}}{\text{1000}\text{mg}}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\\ =35\times {10}^{-6}\text{kg}\end{array}$

Find the volume of the cup in unit of meter-cube.

  $\begin{array}{c}\text{240}\text{mL}\text{=240}\text{mL×}\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{×}\frac{\text{1}{\text{m}}^{\text{3}}}{\text{1000}\text{L}}\\ =2.40\times {10}^{-4}{\text{m}}^{\text{3}}\end{array}$

Next using equation-1 find the concentration in unit of kilogram per meter-cube.

  $\begin{array}{c}\text{Concentration =}\frac{{\text{35×10}}^{\text{-6}}\text{kg}}{\text{2}\text{.40}{\text{×10}}^{\text{-4}}{\text{m}}^{\text{3}}}\\ \text{=}\text{0}\text{.146}{\text{kg/m}}^{\text{3}}\end{array}$

Conclusion:

Thus,the concentration of cholesterol in a cup of milk is $\text{0}{\text{.146 kgm}}^{-3}$ and $\text{0}\text{.146 mg/mL}$.

To determine

(c)

The volume of the cell.

Answer to Problem 49QAP

The volume of the cell is $5.236\times {10}^{-16}{\text{m}}^{\text{3}}$.

Explanation of Solution

Given info:

Diameter of the human cell is $1\text{0}\text{μm}$.

Formula used:

The formula to find the volume of the cell is,

$V_{Cel}{}_l=\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$

Equation-1

Here, $d$ is the diameter of the cell.

Calculation:

Find the volume of the cell by substituting the value of diameter in equation-1.

  $\begin{array}{c}{V}_{cell}=\frac{4}{3}\pi {\left(\frac{10}{5}\text{μm}\right)}^3\\ =523.6{\left(\text{μm}\right)}^{\text{3}}\end{array}$

Next convert the unit form micrometer cube to meter cube.

  $\begin{array}{l}{V}_{cell}=523.6{\left(\text{μm}\right)}^{\text{3}}\times \frac{{\left({\text{10}}^{-6}\text{m}\right)}^{\text{3}}}{1{\left(\text{μm}\right)}^{\text{3}}}\\ =5.236\times {10}^{-16}{\text{m}}^{\text{3}}\end{array}$

Conclusion:

Thus, form equation-2the volume of the cell is $5.236\times {10}^{-16}{\text{m}}^{\text{3}}$.

To determine

(d)

The active ingredient $100$ tablets of low strength aspirin.

Answer to Problem 49QAP

The $100$ tablets of low strength aspirin contains $\text{81×}{\text{10}}^{\text{-4}}\text{kg}$ active ingredient.

Explanation of Solution

Given info:

Each tablet of low strength aspirin contains $\text{81mg}$ of active ingredient.

Introduction:

The International System of Units is the most widely used system of measurement. It has seven basic unit and twenty-two derived units.

Finding the active ingredient of each tablet of low strength aspirin:

Convert the unit of active ingredient intake from milligram to kilogram.

  $\begin{array}{c}\text{81 mg}\text{=81 mg}\times \frac{1\text{kg}}{{10}^6\text{g}}\\ =81\times {10}^{-4}\text{kg}\end{array}$

So, the hundred tablets contain

  $\begin{array}{c}M=81\times {10}^{-6}\text{kg}\times \text{100}\\ =81\times {10}^{-4}\text{kg}\end{array}$

Conclusion:

The $100$ tablets of low strength aspirin contains $\text{81×}{\text{10}}^{\text{-4}}\text{kg}$ active ingredient.

To determine

(e)

The average flow rate of urine from kidney in unit of ${\text{m}}^{\text{3}}\text{/s}$.

Answer to Problem 49QAP

The average flow rate of urine from kidney is $2.08{\text{m}}^{\text{3}}\text{/s}$.

Explanation of Solution

Given info:

The average flow rate of urine from kidney is $1.2\text{mL/min}$.

Introduction:

The International System of Units is the most widely used system of measurement. It has seven basic unit and twenty-two derived units.

Find the volume of unit from milliliter per minute to meter-cube per second.

  $\begin{array}{c}\text{1}\text{.2}\text{mL}\text{=1}\text{.2}\text{mL×}\frac{\text{1}\text{L}}{\text{1000}\text{mL}}\text{×}\frac{\text{1}{\text{m}}^{\text{3}}}{\text{1000}\text{L}}\\ =1.2\times {10}^{-6}{\text{m}}^{\text{3}}\end{array}$

Find the unit of time from minutes to second.

  $\begin{array}{c}\text{1}\text{min}\text{=1}\text{min×}\frac{60\text{s}}{\text{1}\text{min}}\\ =60\text{s}\end{array}$

Thus, the unit of flow rate in meter cube is,

  $\begin{array}{l}\text{Q =}\frac{\text{1}{\text{.2×10}}^{\text{-6}}{\text{m}}^3}{60\text{s}}\\ \text{=}\text{2}\text{.08}{\text{m}}^{\text{3}}\text{/s}\end{array}$

Conclusion:

The average flow rate of urine from kidney is $2.08{\text{m}}^{\text{3}}\text{/s}$.

To determine

(f)

Thedensity of blood protein in unit of ${\text{m}}^{\text{3}}\text{/s}$.

Answer to Problem 49QAP

The density of blood protein is $\text{1}\text{.6}\times {\text{10}}^3{\text{kg/m}}^{\text{3}}$.

Explanation of Solution

Given info:

The density of blood protein is $\text{1}\text{.4}{\text{g/cm}}^3$.

Introduction:

The International System of Units is the most widely used system of measurement. It has seven basic unit and twenty-two derived units.

Find the density of blood protein in unit of ${\text{kg/m}}^{\text{3}}$.

Convert the unit of mass form grams to kilograms

  $\begin{array}{c}\text{1}\text{.4}\text{g}\text{=1}\text{.4}\text{g}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\\ \text{=}\text{1}\text{.4}{\text{×10}}^{\text{-3}}\text{kg}\end{array}$

Find the unit of volume from centi-meter cube to meter cube.

  $\begin{array}{c}\text{1}{\text{cm}}^3\text{=1}{\text{cm}}^3\text{×}\frac{\text{1}{\text{m}}^3}{{\text{10}}^6{\text{cm}}^3}\\ ={\text{10}}^{-6}{\text{m}}^3\end{array}$

Thus, the density in unit of ${\text{kg/m}}^{\text{3}}$ is,

  $\begin{array}{c}\text{Density =}\frac{\text{1}{\text{.2×10}}^{\text{-3}}\text{kg}}{{\text{10}}^{\text{-6}}{\text{m}}^{\text{3}}}\\ \text{=}\text{1}\text{.6}\text{×}{\text{10}}^{\text{3}}{\text{kg/m}}^{\text{3}}\end{array}$

Conclusion:

The density of blood protein is $\text{1}\text{.6}\times {\text{10}}^3{\text{kg/m}}^{\text{3}}$.