Answer:
Stars which never disappear below the horizon are called <u>circumpolar</u> stars.
Answer:
You have a chemical reaction that is occurring
Answer:
Group 17/VIIA
Highly reactive
7 electrons in their outer shell
They become stable and have noble gas configurations when they gain one more electron from metal
Explanation:
halogen, any of the six nonmetallic elements that constitute Group 17 (Group VIIa) of the periodic table. The halogen elements are fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At), and tennessine (Ts).
Due to their high effective nuclear charge, halogens are highly electronegative. Therefore, they are highly reactive and can gain an electron through reaction with other elements.
Halogens are very reactive because they have seven valence electrons and need one more to have eight valence electrons (an octet). They react with metals and other halogens to get an octet. When this happens, the atoms become stable and have noble gas configurations.
Answer:
a) 0.11
b)76.9
c) 8.8
d) 1.7*10^-4
Explanation:
Step 1: Data given
K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 2: Formula of K
aA(g) + bB(g) ⇌ cC(g) + dD(g)
K = [C]^c *[D]^d / [A]^a * [B]^b
K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]
Step 3:
a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
1/2N2 + 3/2H2(g) ⇌ NH3(g) =>K' = 
K' = 
= 0.11
<em>b. 2NH3(g) ⇌ N2(g) + 3H2(g)</em>
N2(g) + 3H2(g) ⇌ 2NH3
2NH3(g) ⇌ N2(g) + 3H2(g) =>K' = 1/K
K' = 1/(1.3*10^-2) = 76.9
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
=>K' = 
K' = 
K' = 8.8
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
2N2(g) + 6H2(g) ⇌ 4NH3(g)
K' = K²
K' = (1.3*10^-2)²
K' = 1.7 *10 ^-4
