Answer:
I'm fairly sure it's D
Explanation:
If they're more spread out than they're roots in theory should have more space. Sorry if I'm wrong
Answer:
The toxicant absorbtion can be reduced after exposure to the skin the surrounding clothing shoes or gloves should be removed or torn off than the part of the body which was exposured to the toxicant should be immediatly washed using clean running water for a while, with cold shower being the most recommended splashing method. In case of absorption for orally consumed chemicals should check on any remaining toxicant residue and be removed from the mouth. Vomiting should be induced to patients that are still conscious by providing them with liquids that can provoke vomiting. This will help in removing the toxicant in the intestinal and reduce their effect. Gastric lavage should then be done to induce diarrhea.
Explanation:
Because displacement of water is the convenient way to obtain gas.
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!