I need help with this question ASAP

Select all that apply. choose the two correct statements below. the elements of the halogen family are very reactive because: th

Harman [31]

Halogens are a group of elements consisting of Fluorine, Chlorine, Bromine, Iodine and Astatine. In their ionic form, they have a superscript of -1, for example, chloride ion is Cl-1. These means that they readily accept one electron in order to achieve the Octet rule. The Octet rule states that each atom must contain 8 electrons in their valence shell for it to be stable. The most stable set of elements are the noble gases. Because they already fulfill the Octet rule, they no longer take part in reactions. Halogens are also very electronegative, meaning, they attract more electrons toward them. This is also a consequence of the Octet rule.

From the choices, the answers would be:
<span>they require only one electron to complete their outer shell
they have a high electronegativity</span>

6 0

3 years ago

What will happen if a peeled banana is put on a hotplate?

inysia [295]

It will melt or the molecules inside of it will get hot

8 0

3 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:

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