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SVETLANKA909090 [29]
3 years ago
14

In an experiment, a student wants to increase the rate of a reaction that involves gases. Which change to the reactants would

Chemistry
1 answer:
Rainbow [258]3 years ago
3 0

Answer:

  • The last option: <u><em>Decrease the volume to increase pressure and to increase concentration.</em></u>

Explanation:

You can support your choice on basis of the collision theory.

According to the collision theory, the chemical reactions happen when the molecules collide with each other, in the correct orientation and with enough kinetic energy to overcome the activation energy.

As consequence, the larger the number of collisions the larger the rate of a reaction.

In the case of a reaction that involves gases, decreasing the volume, will increase the pressure and the concentration (volume is inversely related to both the pressure and the concentration) causing the molecules to be closer to each other and to collide with higher frequency, this is you will expect more collisions, and so an increase on the rate of the reaction. That is expressed by the last choice: decrease the volumen to increase pressure and to increase concentration.

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Read 2 more answers
I NEED HELP PLEASE!!!! CHEMISTRY QUESTION: If 38 g of Li3P and 15 grams of Al2O3 are reacted, what total mass of products will r
maksim [4K]

Answer:

21.5 g.

Explanation:

Hello!

In this case, since the reaction between the given compounds is:

2Li_3P+Al_2O_3\rightarrow 3Li_2O+2AlP

We can see that according to the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction, the total mass of products equals the total mass of reactants based on the stoichiometric proportions; in such a way, we first need to compute the reacted moles of Li3P as shown below:

n_{Li_3P}^{reacted}=38gLi_3P*\frac{1molLi_3P}{51.8gLi_3P}=0.73molLi_3P

Now, the moles of Li3P consumed by 15 g of Al2O3:

n_{Li_3P}^{consumed \ by \ Al_2O_3}=15gAl_2O_3*\frac{1molAl_2O_3}{101.96gAl_2O_3} *\frac{2molLi_3P}{1molAl_2O_3} =0.29molLi_3P

Thus, we infer that just 0.29 moles of 0.73 react to form products; which means that the mass of formed products is:

m_{Li_2O}=0.29molLi_3P*\frac{3molLi_2O}{2molLi_3P} *\frac{29.88gLi_2O}{1molLi_2O} =13gLi_2O\\\\m_{AlP}=0.29molLi_3P*\frac{2molAlP}{2molLi_3P} *\frac{57.95gAlP}{1molAlP} =8.5gAlP

Therefore, the total mass of products is:

m_{products}=13g+8.5g\\\\m_{products}=21.5g

Which is not the same to the reactants (53 g) because there is an excess of Li₃P.

Best Regards!

7 0
3 years ago
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