M + e- ---> M-1 Is M an oxidizing or reducing agent?

Chemistry

1 answer:

Irina-Kira [14]2 years ago

3 0

Answer:

Oxidizing agent

Explanation:

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Given what you observe at the molecular/atomic level, why would surface area have this effect?

LenKa [72]

Answer:

Increasing the surface area of a reactant increases the frequency of collisions and increases the reaction rate. Several smaller particles have more surface area than one large particle. The more surface area that is available for particles to collide, the faster the reaction will occur.

Explanation:

5 0

2 years ago

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:

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