Answer:
- 0.07 °C
Explanation:
At constant pressure and number of moles, Using Charle's law
Given ,
V₁ = 439 mL = 0.439 L ( 1 L = 0.001 mL )
V₂ = 0.378 L
T₁ = 317.15 K
T₂ = ?
Using above equation as:
The conversion of T(K) to T( °C) is shown below:
T( °C) = T(K) - 273.15
So, <u>T = 273.08 - 273.15 °C = - 0.07 °C</u>
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](https://tex.z-dn.net/?f=2Li_3P%2BAl_2O_3%5Crightarrow%203Li_2O%2B2AlP)
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](https://tex.z-dn.net/?f=n_%7BLi_3P%7D%5E%7Breacted%7D%3D38gLi_3P%2A%5Cfrac%7B1molLi_3P%7D%7B51.8gLi_3P%7D%3D0.73molLi_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](https://tex.z-dn.net/?f=n_%7BLi_3P%7D%5E%7Bconsumed%20%5C%20by%20%5C%20Al_2O_3%7D%3D15gAl_2O_3%2A%5Cfrac%7B1molAl_2O_3%7D%7B101.96gAl_2O_3%7D%20%2A%5Cfrac%7B2molLi_3P%7D%7B1molAl_2O_3%7D%20%3D0.29molLi_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](https://tex.z-dn.net/?f=m_%7BLi_2O%7D%3D0.29molLi_3P%2A%5Cfrac%7B3molLi_2O%7D%7B2molLi_3P%7D%20%2A%5Cfrac%7B29.88gLi_2O%7D%7B1molLi_2O%7D%20%3D13gLi_2O%5C%5C%5C%5Cm_%7BAlP%7D%3D0.29molLi_3P%2A%5Cfrac%7B2molAlP%7D%7B2molLi_3P%7D%20%2A%5Cfrac%7B57.95gAlP%7D%7B1molAlP%7D%20%3D8.5gAlP)
Therefore, the total mass of products is:
![m_{products}=13g+8.5g\\\\m_{products}=21.5g](https://tex.z-dn.net/?f=m_%7Bproducts%7D%3D13g%2B8.5g%5C%5C%5C%5Cm_%7Bproducts%7D%3D21.5g)
Which is not the same to the reactants (53 g) because there is an excess of Li₃P.
Best Regards!
<span>In the electron cloud model, the denser areas represent that there is a great probability that a good number of electrons are ganged up or crowded in that area. The electrons affect the density of some parts of the electron cloud when they condense in those locations.</span>