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3 years ago

Onsider this reversible reaction:

Chemistry

2 answers:

son4ous [18]3 years ago

5 0

You can shift the equilibrium to the right if you

• Increase the concentration of PCl_5

• Increase the temperature

• Decrease the concentration of Cl_2

• Decrease the total pressure of the system

Decreasing the concentration of PCl_3 will shift the equilibrium to shift to the right, but there will be less PCl_3 at the new equilibrium.

solong [7]3 years ago

3 0

Answer:

PCl₅(g) + heat ↔ PCl₃(g) + Cl₂(g)

On increasing the concentration of PCl5(g), the equilibrium will shift in forward direction and produce more PCl₃.

Explanation:

According to Le Chatlier's Principle " If we apply any external stress on a system at equilibrium, the system minimize the effect by itself to regain equilibrium". In above reaction initially the system is in equilibrium. when we increase the concentration of reactant (PCl₅), the system goes to forward and produce more product (PCl₃).

After some time reaction will go in backward direction to regain equilibrium position while consuming PCl₃ and Cl₂. Finally reaction will again in equilibrium position.

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Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2

Elena-2011 [213]

Answer:

Approximately $0.224\;\rm L$ , assuming that this reaction took place under standard temperature and pressure, and that $\rm CO_2$ behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that $\rm CaCO_3$ is the limiting reactant. How many moles of $\rm CO_2$ would be produced?

Look up the relative atomic mass of $\rm Ca$ , $\rm C$ , and $\rm O$ on a modern periodic table:

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

Calculate the formula mass of $\rm CaCO_3$ :

$\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of formula units in $1\; \rm g$ of $\rm CaCO_3$ using its formula mass:

$\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}$ .

In the balanced chemical equation, the ratio between the coefficient of $\rm CaCO_3$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1$ .

In other words, for each mole of $\rm CaCO_3$ formula units consumed, one mole of $\rm CO_2$ would be produced.

If $\rm CaCO_3$ is indeed the limiting reactant, all that approximately $1.00\times 10^{-2}\; \rm mol$ of $\rm CaCO_3\!$ formula would be consumed. That would produce approximately $1.00\times 10^{-2}\; \rm mol\!$ of $\rm CO_2$ .

On the other hand, assume that $\rm HCl$ is the limiting reactant.

Convert the volume of $\rm HCl$ to $\rm dm^{3}$ (so as to match the unit of concentration.)

$\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $5.00\times 10^{-2}\; \rm dm^{3}$ of this $\rm 0.05\; \rm mol \cdot dm^{-3}$

$\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}$ .

Notice that in the balanced chemical reaction, the ratio between the coefficient of $\rm HCl$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}$ .

In other words, each mole of $\rm HCl$ molecules consumed would produce only $0.5\;\rm mol$ of $\rm CO_2$ molecules.

Therefore, if $\rm HCl$ is the limiting reactant, that $2.50 \times 10^{-3}\; \rm mol$ of $\rm HCl\!$ molecules would produce only one-half as many (that is, $1.25\times 10^{-3}\; \rm mol$ ) of $\rm CO_2$ molecules.

If $\rm CaCO_3$ is the limiting reactant, $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. However, if $\rm HCl$ is the limiting reactant, $1.25\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be produced.

In reality, no more than $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. The reason is that all $\rm CaCO_3$ would have been consumed before $\rm HCl$ was.

After finding the limiting reactant, approximate the volume of the $\rm CO_2\!$ produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately $22.4\; \rm L$ .

If $\rm CO_2$ behaves like an ideal gas, the volume of that $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be approximately $\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L$ .

3 0

3 years ago

Dominic made the table below to organize his notes about mixtures,

Troyanec [42]

Answer: Although it is possible to have more than one state, it is also possible to have only one state.

Explanation: Just took the test

6 0

3 years ago

Calculate the pH of a solution that has [H3O+]=4.3 x 10^-5 M

Neko [114]

Answer:

pH = 4.4

Explanation:

pH = -log[H₃O⁺]

= -log(4.3 × 10⁻⁵)

pH = 4.37 = 4.4

I'm actually learning this in my chemistry class right now lol. Hope this helps though. :)

6 0

3 years ago

Which atom is most lickley to be shiny?<br> Sulfur<br> Boron<br> Calcium <br> Carbon

Arisa [49]

Answer:

Calcium

Explanation: Calcium is a fairly soft metal with a shiny silver surface when first cut.

6 0

4 years ago

A double-replacement reaction takes place when aqueous K2SO4 reacts with aqueous Pb(NO3)2. You would expect one of the products

melomori [17]

A double displacement reaction is one in which the positive and negative ion in one reactant switch places with the positive and negative ions of the second reactant forming two new products.

In general if we have two reactants AB and CD, the double displacement reaction can form the following products:

A⁺B⁻ + C⁺D⁻ → C⁺B⁻ + A⁺D⁻

In the given example:

K₂SO₄ + Pb(NO₃)₂ → PbSO₄ + 2KNO₃

Ans: B)

One of the products would be PbSO₄

5 0

3 years ago

Read 2 more answers