Explanation:
Experiment Initial [CS2] (mol/L) Initial Rate (mol/L·s)
1 0.100 2.7 × 10−7
2 0.080 2.2 × 10−7
3 0.055 1.5 × 10−7
4 0.044 1.2 × 10−7
a) Choose the rate law for the decomposition of CS2.
Comparing equations 1 and 3, reducing the initial concentration by almost half (from 0.100 to 0.055) leads too the rate of reaction to be reduced by almost half (from 2.7 × 10−7 to 1.5 × 10−7).
This signifies that the reaction is a first order reaction.
Rate = k [CS2]
(b) Calculate the average value of the rate constant.
Taking equation 1.
Rate = k [CS2]
k = Rate / [CS2]
k = 0.100 / (2.7 × 10−7) = 0.037 x 10^8 = 3.7 x 10^6s-1
Answer:

Explanation:
Atomic radius is the measurement from the nucleus to the outer edge of the electron cloud.
As you go down a group (vertically) the atomic radius increases because more electron shells are added. As you go across a period horizontally, the atomic radius decreases.
If we look at the halogens group (17), we see they follow this order from top to bottom:
F - Fluorine
Cl - Chlorine
Br - Bromine
I - Iodine
Since it increases down the group, iodine must have the largest atomic radius.
Answer:
Aromatic Hydrocarbons
Explanation:
Aromatic (Pleasant Odour) Hydrocarbons are those having pleasant odours.
Answer:
32
Step-by-step explanation:
There are two ways you can count the valence electrons.
A. From the Periodic Table
1 × P (Group 15) = 5
4 × O (Group 16) = 4 × 6 = 24
+3 e⁻ (for the charges) = <u> 3</u>
Total = 32
B. From the Lewis structure
In the <em>Lewis structure</em> (below), each line (bond) represents a pair of bonding electrons, and each dot represents an unbound electron (half a lone pair).
5 lines (bonds) = 5 × 2 = 10
3 single-bonded O atoms = 3 × 6 = 18
1 double-bonded O atom = <u> 4</u>
Total = 32
Answer:
2,4,2
Explanation:
by looking at the subscripts (small number behind the element) you can tell how many of these atoms are presented in the compound.
your compound is C2H4O2
so there are two carbon atoms present, four hydrogen atoms present, and 2 oxygen atoms present.
hope this helps :)