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Zigmanuir [339]

3 years ago

15

According to Wookieepedia, Kylo Ren stands 1.89 m tall and his mass is 89 kg. Calculate his earth-weight in lbm. (2 pts) a. b. A

ssuming he has a 50% waist to height ratio, calculate his diameter in inches.

1 answer:

harina [27]3 years ago

6 0

Answer:

a) 6312.12 lbm

b) D = 11.843 inch

Explanation:

GIVEN DATA:

Height of person = 1.89 m = 74.41 inch

mass of person m = 89 kg = 196.211 lbm

a) person earth weight =mg

where g is acceleration due to gravity = 32.17 ft/sec^2

$= 196.211\times 32.17 = 6312.12 lbm$

b) waist to height ratio = 0.5

$waist\ to\ height\ ratio = \frac{circumference\ of\ the\ waist}{height\ of\ the\ person}$

circumference of waist = $height\ of\ person \times waist\ to\ height ratio$

$= 74.41 inch \times 0.5$

=37.21 inch

we know that circumference is given as $= 2\pi r\ or\ \pi D= 37.21$

D = 11.843 inch

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Balanced equation for this reaction:

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Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

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Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

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$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

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