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Andrews [41]
3 years ago
13

Example. A Student wants to increase the maximum if sugar that can dissolve in water..She crushes the sugar and then stirs it in

to the water .Does This work?
Chemistry
2 answers:
Advocard [28]3 years ago
8 0

Answer:

Yes this works.

Explanation:

This works because when the sugar is crushed, more of the surface area of the sugar gets exposed to the solvent, allowing the solvent to dissolve it a lot quicker than when it wasn't crushed.

Stirring also helps the sugar particles dissolve faster. This is because the particles are now agitated, which increases their kinetic energy.  This increase in Kinetic energy is reflected by increase in temperature, which helps the particles dissolve faster.

Delvig [45]3 years ago
6 0

Answer: yes it does work.

Explanation:

When the sugar is crushed by the student, an increase in the number of molecules of sugar occurs. Meaning that more number of the solute (sugar molecules) are able to interact with the solvent (the water).

Therefore, there will be more number of collisions taking place between the solute and solvent molecules.

Hence, we can say it works because this leads to more interaction or collisions between the solute and solvent particles, as a result of this increased interaction between solute and the solvent, sugar will dissolve readily into the water.

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What change would shift the equilibrium system to the left?
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Answer:

Adding more of gas C to the system

Explanation:

  • <em>Le Châtelier's principle</em><em> states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>

1) Adding more of gas C to the system:

Adding more C gas will increase the concentration of the products side. So, the reaction will be shifted to the left to attain the equilibrium again.

2) Heating the system:

Heating the system will increase the concentration of the reactants side as the reaction is endothermic. so, the reaction will be shifted to the right to attain the equilibrium again.

3) Increasing the volume:

has no effect since the no. of moles of gases is the same in both reactants and products sides.

4) Removing some of gas C from the system:

Removing some of gas C from the system will decrease the concentration of the products side. So, the reaction will be shifted to the right to attain the equilibrium again.

<em>So, the right choice is: Adding more of gas C to the system.</em>

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Ozone, o3, is a product in automobile exhaust by the reaction represented by the equation no2(g) + o2(g) --&gt; 3no(g) + o3(g).
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Answer: 2.1 g mass of ozone(O_{3}) is predicted to form from the reaction of 2.0 g NO_{2} in a car's exhaust and excess oxygen

Given information : Mass of NO_{2} = 2.0 g and O_{2} is in excess.

We need to calculate the mass of ozone (O_{3})

Mass of ozone(O_{3}) is calculated with the help of mass of NO_{2} using stoichiometry.

NO_{2} + O_{2}\rightarrow NO + O_{3}

Step 1 : Convert grams of NO_{2} to moles of NO_{2}.

Moles = \frac{Grams}{Molar mass}

Molar mass of NO_{2} = 46.0 g/mol

Moles = \frac{2.0g}{46.0\frac{g}{mol}}

Moles of NO_{2} = 0.043 mol

Step 2 : Find the moles of O_{3} using moles of NO_{2}.

Moles of O_{3} is calculated by using moles of NO_{2} with the help of mole ratio.

A mole ratio is ​the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.

From the balanced chemical equation we can see that coefficient of NO_{2} is 1 and coefficient of O3 is 1 , so mole ratio of O_{3} to NO_{2} is 1:1

Moles of O_{3} = (0.043 mol NO_{2})\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}

Moles of O_{3} = (0.043)\times \frac{(1 mol O_{3})}{(1)}

Moles of O_{3} = 0.043 mol

Step 3 : Convert moles of O_{3} to grams of O_{3}

Grams = Moles X Molar mass

Molar mass of O_{3} = 48.0 g/mol

Grams = (0.043 mol O_{3})\times (\frac{48 g O_{3}}{1 mol O_{3}})

Grams = (0.043)\times (\frac{48 g O_{3}}{1})

Grams = 2.1 g O_{3}

Note : The above three steps can also be done using a single step setup.

Grams of O_{3} = (2.0 gNO_{2})\times \frac{(1mol NO_{2})}{(46.0 g NO_{2})}\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}\times \frac{(48.0 g O_{3})}{(1 mol O_{3})}

Grams of O_{3} = (2.0 )\times \frac{(1)}{(46.0 )}\times \frac{(1)}{(1)}\times \frac{(48.0 g O_{3})}{(1)}

Grams of O_{3} = 2.1 grams


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