Chapter 1, Problem 48R

(a)

To determine

ToConvert: $1500\text{mL}$ to $\text{L}$ .

Answer to Problem 48R

  $1500\text{mL=1}\text{.5}\text{L}$ .

Explanation of Solution

Formula:

  $\text{1}\text{L=1000}\text{mL}$  ...... (1)

Calculation:

Given $1500\text{mL}$ , this can be written as $\text{1000}\text{mL+500}\text{mL}$

  $\text{1000}\text{mL+500}\text{mL=}\text{1}\text{.5}\text{L}$

Conclusion:

Hence, the required answer is $1500\text{mL=1}\text{.5}\text{L}$ .

(b)

To determine

To: Convert $2\text{km}$ to $\text{cm}$ .

Answer to Problem 48R

  $2000\text{cm}$ .

Explanation of Solution

Formula:

  $\text{1}\text{km}\text{=1000}\text{cm}$  ...... (1)

Calculation:

Given $2\text{km}$ , this can be written as $\text{1}\text{km + 1}\text{km}$

  $\begin{array}{l}=1\text{km}\text{+}\text{1}\text{km}\\ \text{=1000}\text{cm+1000}\text{cm}\\ \text{=2000}\text{cm}\end{array}$

Conclusion:

Hence, the required answer is $2000\text{cm}$ .

(c)

To determine

To: Convert $5.8\text{dg}$ to $\text{mg}$ .

Answer to Problem 48R

  $580\text{gm}$ .

Explanation of Solution

Formula:

  $\text{1}\text{dg}\text{=}\text{0}\text{.1}\text{g}$  ...... (1)

  $\text{1}\text{g}\text{=}\text{1000}\text{mg}$  ...... (2)

Calculation:

Given $5.8\text{dg}$ ,

Used the formula (1) and (2)

  $\begin{array}{c}5.\text{8}\text{dg}\text{=}\text{0}\text{.1×5}\text{.8}\\ \text{=}\text{0}\text{.58}\text{g}\\ \text{=}\text{1000×}\text{0}\text{.58}\\ \text{=}\text{580}\text{mg}\end{array}$

Conclusion:

Hence, the required answer is $580\text{gm}$ .

(d)

To determine

To: Convert ${22}^{\circ }\text{C}$ to $\text{K}$ .

Answer to Problem 48R

  $295.15\text{K}$ .

Explanation of Solution

Formula:

Formula for converting ${}^{\circ }\text{C}$ to $\text{K}$ is

  ${\text{0}}^{\text{o}}\text{C+273}\text{.15}\text{K}$  ...... (1)

Calculation:

Given ${22}^{\text{o}}\text{C}$ ,

Substitute the value ${22}^{\text{o}}\text{C}$ in place of $0^{\text{o}}\text{C}$ in equation (1)

  $\begin{array}{c}{\text{22}}^{\circ }\text{C =}{\text{22}}^{\text{o}}\text{C+273}\text{.15}\text{K}\\ \text{=295}\text{.15K}\end{array}$

Conclusion:

Hence, the required answer is $295.15\text{K}$ .