Chapter 2.7, Problem 4PPC

(a)

Interpretation Introduction

Interpretation:

The mass and number of plain and jam-filled doughnuts to be calculated.

Concept Introduction:

Density: the density of the substance is calculated using the formula:

$\text{density = }\frac{\text{mass}}{\text{volume}}$

Atoms: Atoms consist of tiny particles called protons, neutrons and electrons. Proton and neutrons are present in the nucleus and the electron resides around the nucleus. The protons number will be same as the electrons count in the atom.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Answer to Problem 4PPC

The mass of plain and jam-filled doughnuts is $\text{40g and 480g respectively}$.

Explanation of Solution

Calculate the mass of dozen plain and jam-filled doughnuts:

$\begin{array}{l}\text{Given: }\\ \text{Average Mass of plain doughnuts : 32}\text{.6 g}\text{.}\\ \text{Average Mass of jam-filled doughnuts : 40}\text{.0 g}\text{.}\\ \\ \text{Mass}\text{of}\text{dozen}\text{doughnuts =}\text{12(32}\text{.6) =}391.2g.\\ \text{Mass}\text{of}\text{dozen}\text{jam-filled}\text{doughnuts =}\text{12(40}\text{.0) =}480g.\\ \end{array}$

The mass ofplainand jam-filled doughnuts is $\text{40g and 480g respectively}$, as per the calculation done above.

Conclusion

The mass and number of plain and jam-filled doughnuts was calculated.

(b)

Interpretation Introduction

Interpretation:

The mass and number of plain and jam-filled doughnuts to be calculated.

Concept Introduction:

Density: the density of the substance is calculated using the formula:

$\text{density = }\frac{\text{mass}}{\text{volume}}$

Atoms: Atoms consist of tiny particles called protons, neutrons and electrons. Proton and neutrons are present in the nucleus and the electron resides around the nucleus. The protons number will be same as the electrons count in the atom.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Answer to Problem 4PPC

The Number of plain and jam-filled doughnuts is $\text{30}\text{.7, 25 respectively}$,

Explanation of Solution

Calculate the Number of plain and jam-filled doughnuts:

$\begin{array}{l}\text{Given: }\\ \text{Average Mass of plain doughnuts : 32}\text{.6 g}\text{.}\\ \text{Average Mass of jam-filled doughnuts : 40}\text{.0 g}\text{.}\\ \\ \underset{\_}{\text{Calculate}\text{Number of}\text{plain}\text{doughnuts in}\text{1}\text{Kg}\text{:}}\\ \text{Number of}\text{doughnuts =}\text{x}\\ \text{1}\text{Kg}\text{=}\text{no}\text{.}\text{of doughnuts}\text{(mass of plain doughnuts)}\\ \text{1000 g =}\text{x }\text{(32}\text{.6g)}\\ \text{x = }\frac{\text{1000}}{\text{32}\text{.6}}\text{ = 30}\text{.7 doughnuts}\text{.}\\ \boxed{Numberofdoughnutsin1Kgis30.7}\\ \\ \underset{\_}{\text{Calculate}\text{Number of}\text{jam-filled}\text{doughnuts in}\text{1}\text{Kg}\text{:}}\\ \text{Number of}\text{jam-filled}\text{doughnuts =}\text{y}\\ \text{1}\text{Kg}\text{=}\text{no}\text{.}\text{of}\text{jam-filled doughnuts}\text{(mass of plain doughnuts)}\\ \text{1000 g =}\text{y }\text{(40}\text{.0g)}\\ \text{y = }\frac{\text{1000}}{\text{40}\text{.0}}\text{ = 25 jam-filled doughnuts}\text{.}\\ \boxed{\text{Number of}\text{jam-filled}\text{doughnuts in}\text{1}\text{Kg }\text{is 25}}\text{.}\end{array}$

The Number of plainand jam-filled doughnuts is $\text{30}\text{.7, 25 respectively}$, as per the calculation done above.

Conclusion

The mass and number of plain and jam-filled doughnuts was calculated.

(c)

Interpretation Introduction

Interpretation:

The mass and number of plain and jam-filled doughnuts to be calculated.

Concept Introduction:

Density: the density of the substance is calculated using the formula:

$\text{density = }\frac{\text{mass}}{\text{volume}}$

Atoms: Atoms consist of tiny particles called protons, neutrons and electrons. Proton and neutrons are present in the nucleus and the electron resides around the nucleus. The protons number will be same as the electrons count in the atom.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Answer to Problem 4PPC

The total mass of plain doughnuts is $\text{815}\text{g}$

Explanation of Solution

Calculate the mass of plain doughnuts:

$\begin{array}{l}\text{Given: }\\ \text{Average Mass of plain doughnuts : 32}\text{.6 g}\text{.}\\ \text{Average Mass of jam-filled doughnuts : 40}\text{.0 g}\text{.}\\ \\ \underset{\_}{\text{Calculate}\text{the}\text{mass}\text{of}\text{plain}\text{doughnuts:}}\\ \text{Given : Mass}\text{of}\text{plain}\text{doughnuts =}\text{No}\text{.of}\text{1}\text{Kg}\text{of}\text{jam-filled}\text{doughnuts}\text{.}\\ \text{thus,}\\ \text{Mass}\text{of}\text{plain}\text{doughnuts}\text{=}\text{25}\text{(32}\text{.6 g)}\text{=}\text{815}\text{g}\text{.}\\ \boxed{\text{Mass}\text{of}\text{plain}\text{doughnuts =}\text{815}\text{g}\text{.}}\\ \end{array}$

The total mass of plain doughnuts is $\text{815}\text{g}$ as per the calculation done above.

Conclusion

The mass and number of plain and jam-filled doughnuts was calculated.

(d)

Interpretation Introduction

Interpretation:

The mass and number of plain and jam-filled doughnuts to be calculated.

Concept Introduction:

Density: the density of the substance is calculated using the formula:

$\text{density = }\frac{\text{mass}}{\text{volume}}$

Atoms: Atoms consist of tiny particles called protons, neutrons and electrons. Proton and neutrons are present in the nucleus and the electron resides around the nucleus. The protons number will be same as the electrons count in the atom.

$\begin{array}{l}{\text{The element symbol : }}_{\text{Z}}^{\text{A}}\text{X,}\\ \text{where, A (mass number) = no}\text{.of protons + no}\text{.of neutrons}\text{.}\\ \text{ Z (atomic number) = no}\text{. of protons}\text{. (electrons = protons)}\text{.}\end{array}$

Answer to Problem 4PPC

The total mass of dozen doughnuts is $\text{413}\text{.3g}$

Explanation of Solution

Calculate total mass of dozen doughnuts:

$\begin{array}{l}\text{Given: }\\ \text{Average Mass of plain doughnuts : 32}\text{.6 g}\text{.}\\ \text{Average Mass of jam-filled doughnuts : 40}\text{.0 g}\text{.}\\ \\ \underset{\_}{\text{Calculate}\text{the}\text{total}\text{mass}\text{of}\text{dozen}\text{doughnuts:}}\\ \text{Given : Total Mass}\text{of}\text{dozen}\text{doughnuts =}\text{3 (Jam-filled doughnuts) + 9 (plain doughnuts)}\text{.}\\ \text{thus,}\\ \text{Total Mass}\text{of}\text{dozen}\text{doughnuts}\text{=}3(40.0g)+\text{9}\text{(32}\text{.6 g)}\text{=}\text{413}\text{.4}\text{g}\text{.}\\ \boxed{\text{Total Mass}\text{of}\text{dozen}\text{doughnuts =}\text{413}\text{.4}\text{g}\text{.}}\\ \end{array}$

The total mass of dozen doughnuts is $\text{413}\text{.3g}$ as per the calculation done above.

Conclusion

The mass and number of plain and jam-filled doughnuts was calculated.