Answer:
A binary covalent compound is composed of two different elements (usually nonmetals). For example, a molecule of chlorine trifluoride, ClF3 contains 1 atom of chlorine and 3 atoms of fluorine.
Rule 1. The element with the lower group number is written first in the name; the element with the higher group number is written second in the name. Exception: when the compound contains oxygen and a halogen, the name of the halogen is the first word in the name.
Rule 2. If both elements are in the same group, the element with the higher period number is written first in the name.
Rule 3. The second element in the name is named as if it were an anion, i.e., by adding the suffix -ide to the root of the element name (e.g., fluorine = F, "fluoride" = F-; sulfur = S, "sulfide" = S2-).
Rule 4. Greek prefixes are used to indicate the number of atoms of each element in the chemical formula for the compound. Exception: if the compound contains one atom of the element that is written first in the name, the prefix "mono-" is not used.
Explanation:
Answer:
earth has a size of 6,371 km wich is greater than mars size 3,389.5 km
Explanation:
Answer:
The answer is 2
The maximum number an subshell can have is 2
Answer:
[H2] = 0.0692 M
[I2] = 0.182 M
[HI] = 0.826 M
Explanation:
Step 1: Data given
Kc = 54.3 at 430 °C
Number of moles hydrogen = 0.714 moles
Number of moles iodine = 0.984 moles
Number of moles HI = 0.886 moles
Volume = 2.40 L
Step 2: The balanced equation
H2 + I2 → 2HI
Step 3: Calculate Q
If we know Q, we know in what direction the reaction will go
Q = [HI]² / [I2][H2]
Q= [n(HI) / V]² /[n(H2)/V][n(I2)/V]
Q =(n(HI)²) /(nH2 *nI2)
Q = 0.886²/(0.714*0.984)
Q =1.117
Q<Kc This means the reaction goes to the right (side of products)
Step 2: Calculate moles at equilibrium
For 1 mol H2 we need 1 mol I2 to produce 2 moles of HI
Moles H2 = 0.714 - X
Moles I2 = 0.984 -X
Moles HI = 0.886 + 2X
Step 3: Define Kc
Kc = [HI]² / [I2][H2]
Kc = [n(HI) / V]² /[n(H2)/V][n(I2)/V]
Kc =(n(HI)²) /(nH2 *nI2)
KC = 54.3 = (0.886+2X)² /((0.714 - X)*(0.984 -X))
X = 0.548
Step 4: Calculate concentrations at the equilibrium
[H2] = (0.714-0.548) / 2.40 = 0.0692 M
[I2] = (0.984 - 0.548) / 2.40 = 0.182 M
[HI] = (0.886+2*0.548) /2.40 = 0.826 M
