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kramer
3 years ago
14

Question 8 helpppppp

Chemistry
2 answers:
Strike441 [17]3 years ago
6 0
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2><u>LE CHATELIER PRINCIPLE</u>:</h2>

Le Chatelier's principle is a principle which predicts the effect of change in condition on a the position of a equilibrium.  

Change could be in Temperature, Moles(Concentration), Pressure.

-------------------------------------------------------------------------------------------------------------

<h2>EFFECTS ON EQUILIBRIUM</h2><h3 /><h3>__________________________________________________</h3><h3>Temperature: </h3>

We could change the temperature by increasing or decreasing it, and there are two types of reactions, endothermic or exothermic, both have different outcomes by changing the condition.

__________________________________________________________

In Endothermic,

When we increase the temperature, It will favor the formation of product thus MORE PRODUCT will be formed - equilibrium will shift towards right i.e. product side.

When we decrease the temperature, It will favor the formation of reactants thus MORE reactant will be formed - equilibrium will shift towards left i.e. reactants side.

__________________________________________________________

In Exothermic,

When we increase the temperature, It will favor the formation of reactants thus MORE REACTANT will be formed - equilibrium will shift towards left i.e. reactants side.

When we decrease the temperature, It will favor the formation of product thus MORE PRODUCT will be formed - equilibrium will shift towards the right i.e. product side

<h2>_____________________________________</h2><h3>Pressure:</h3>

When we increase the pressure, The equilibrium shifts towards the fewer moles of a gas. Wherever there are less moles, it could be any, reactant or product, the increase of Pressure will favor in formation of them.

When we decrease the pressure, The equilibrium shifts towards the more moles of a gas. So whether reactants or product have the higher moles, the decrease in pressure will favor in the formation of them.

<h2>_____________________________________</h2><h3>Concentration of moles:</h3>

When we increase the concentration of moles of any substance, The equilibrium will shift away from that substance. So if we want to make more product and there are more moles of reactant present, then Increase the concentration of reactants.

When we decrease the concentration of any substance, The equilibrium will shift towards that substance. So if we want to make more product, decrease its concentration or add more reactant.

<h2>_____________________________________</h2><h3>Catalyst:</h3>

By the use of the catalyst, There is no effect on the equilibrium. The rate of  both forward and backward reaction are increased with the equal amount.

<h2>_____________________________________</h2><h2>Question:</h2>

By keeping that in mind,

A) Extra NH_{3} will shift the equilibrium towards left i.e. more reactants will be formed.

B) Extra N2 is removed, the equilibrium will shift towards the left i.e. more reactants will be formed.

C) N2 is added, the equilibrium will shift towards right i.e. more PRODUCT will be formed. (I assumed there was a typing error and it meant Addition instead of removal)

D) When we decrease the pressure the equilibrium will shift towards the left i.e. more reactant will be formed.

E) Catalyst has no effect on the system

<h2>_____________________________________</h2><h2>Best Regards</h2><h2>'Borz'</h2><h2 />
solniwko [45]3 years ago
5 0

Answer:

Explanation:

This is lechatlier's principle, which basically talks about how adding or taking away things will disrupt equilibrium and the system will act a certain way to restore equilibrium

1. adding NH3 causes the equilibrium to shift left - produce more reactants

2. removing N2 causes the equilibrium to shift left - produce more reactants

3.  same as number 2

4. decreasing pressure causes equilbrium to shift to the side with more moles so it shifts left and produces more reactants

5. catalyst doesn't change anything about equlibrium, it only affects the reaction's rate

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A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What
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Answer:

Approximately 1.854\; \rm mol\cdot L^{-1}.

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of \rm Sr, \rm O, and \rm H on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

  • \rm Sr: 87.62.
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  • \rm H: 1.008.

Calculate the formula mass of \rm Sr(OH)_2:

M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}.

<h3>Number of moles of strontium hydroxide in the solution</h3>

M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1} means that each mole of \rm Sr(OH)_2 formula units have a mass of 121.634\; \rm g.

The question states that there are 10.60\; \rm g of \rm Sr(OH)_2 in this solution.

How many moles of \rm Sr(OH)_2 formula units would that be?

\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}.

<h3>Molarity of this strontium hydroxide solution</h3>

There are 8.71467\times 10^{-2}\; \rm mol of \rm Sr(OH)_2 formula units in this 47\; \rm mL solution. Convert the unit of volume to liter:

V = 47\; \rm mL = 0.047\; \rm L.

The molarity of a solution measures its molar concentration. For this solution:

\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}.

(Rounded to four significant figures.)

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