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2 years ago

To illustrate the arrangements of particles in a liquid , a teacher instructed 20 students

Chemistry

1 answer:

Monica [59]2 years ago

3 0

Answer:

instructed 20 students to do what

Explanation:

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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

Water (H2O) is made up of molecules with a bent structure, whereas hydrogen sulfide (H2S) is also made up of molecules with a be

jeka94

Śhüt ûp and go pay attention in your class

3 0

3 years ago

Consider the following system at equilibrium at 723 K: 2 NH3 (g) 26.6 kcal N2 (g) 3 H2 (g) Indicate whether each individual chan

ZanzabumX [31]

Answer:

- To increase the temperature as it is a reactant in terms of its endothermicity.

- To remove it will enable more space for the reactant to favor its production.

- To add more reactant in order to increase its equilibrium concentration.

Explanation:

Hello,

The undergoing chemical reaction is:

$2NH_3(g)\rightleftharpoons 3H_2(g)+N_2(g)$

Thus, in order to intensify the amount of nitrogen as the chemical reaction is endothermic, considering the Le Chatelier's principle we state:

- To increase the temperature as it is a reactant in terms of its endothermicity.

- To remove it will enable more space for the reactant to favor its production.

- To add more reactant in order to increase its equilibrium concentration.

Best regards.

4 0

3 years ago

When water freezes it's density gets lower the change in density is different than that of most substances most substances get d

icang [17]

When water is in liquid form its molecules are free to move around.
Water molecules are packed reasonably close together. However when water freezes its molecules take up a hexagonal lattice (repeating structure) which has space in the middle of it.

This is largely due to hydrogen bonding between water molecules (complicated).

As a result water molecules in ice aren't packed as closely together as they are in liquid water so the density of ice is lower than that of liquid water.

Hope that helps. I doubt you need to know about hydrogen bonding.