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3 years ago

In the absence of air a penny and a feather that are dropped from the same height at the same time will

Chemistry

1 answer:

erik [133]3 years ago

6 0

Have the same acceleration/reach the ground at the same time.

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Carbon disulfide burns in oxygen to yield car- bon dioxide and sulfur dioxide according to the chemical equation cs2(l) 3 o2(g)

vampirchik [111]

Answer: Oxygen is the limiting reagent.

Explanation: $CS_2(l)+3O_2(g)\rightarrow CO_2(g)+2SO_2(g)$

As can be seen from the given balanced equation:

3 moles of $O_2$ reacts with 1 mole of $CS_2$

1.52 moles of $O_2$ reacts with= $\frac{1}{3}\times 1.52=0.51moles$ of $CS_2$

Thus $O_2$ is the limiting reagent as it limits the formation of products. (0.91-0.51)= 0.40 moles of $CS_2$ will remain as such and thus $CS_2$ is an excess reagent.

6 0

3 years ago

What is the minimum (non-zero) thickness of a benzene (n = 1.501) thin film that will result in constructive interference when v

ad-work [718]

Explanation:

At each reflecting surface (benzene and glass) there will be 180 degree phase change.

Now, for constructive interference the optical path in benzene is $\lambda$ .

Formula to calculate thickness of a benzene thin film is as follows.

Optical path length through benzene ( $\lambda$ ) = $2 \times n \times d$

Hence, substituting the given values into the above formula as follows.

Optical path length through benzene = $2 \times n \times d$

d = $\frac{\text{Optical path length through benzene}}{2 \times n}$

= $\frac{\lambda}{2 \times n}$

= $\frac{615 \times 10^{-9}}{2 \times 1.501}$ (as 1 nm = $10^{-9}m$

= 204.9 m

Thus, we can conclude that minimum thickness of benzene is 204.9 m.

4 0

3 years ago

10095 m/sec to mile/ sec. 1 km = 0.6214 mile

Anarel [89]

Answer:

Explanation:

to convert any value in miles per second to meters per second, just multiply the value in miles per second by the conversion factor 1609.344. So, 10095 miles per second times 1609.344 is equal to 1.625 × 107 meters per second.

4 0

4 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

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

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

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

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

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

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

6 0

3 years ago

Why check dams can increase the groundwaater level in a region

valkas [14]

Check dams shouldn't be more than 2-3 feet high. The center of it should be at least six inches lower than the edges. They might kill grass linings in channels if water stays high or sediment load is great. That criteria induces a weir effect, which makes water surface levels go upstream for some, if not all flow conditions.
Hope this helps! :)

8 0

4 years ago