Answer:
AuCl
Explanation:
Given parameters:
Mass of Gold = 2.6444g
Mass of Chlorine = 0.476g
Unknown:
Empirical formula = ?
Solution:
Empirical formula is the simplest formula of a compound. Here is the way of determining this formula.
Elements Au Cl
Mass 2.6444 0.476
Molar mass 197 35.5
Number of moles 2.6444/197 0.476/35.5
0.013 0.013
Divide by the
smallest 0.013/0.013 0.013/0.013
1 1
The empirical formula of the compound is AuCl
Answer:
The correct option is: Carbonate ion < Carbon dioxide < Carbon monoxide
Explanation:
Bond energy is defined as the average energy needed to break a chemical covalent bond and signifies the strength of chemical covalent bond.
The bond strength of a covalent bond depends upon the <u>bond length and the bond order.</u>
Carbon monoxide molecule (CO) has two covalent bond and one dative bond. Bond order 2.6
Carbon dioxide (CO₂) has two carbon-oxygen (C-O) double bonds of equal length. Bond order 2.0
Carbonate ion (CO₃²⁻) has three C-O partial double bonds. Bond order 1.5
Also, the bond length is <u>inversely proportional to the bond order and bond strength.</u>
Therefore, <u>order of C-O bond length:</u> Carbon monoxide<Carbon dioxide<Carbonate ion
<u>Order of C-O bond order</u>: Carbonate ion<Carbon dioxide<Carbon monoxide
<u>Order of C-O bond strength or energy</u><u>: Carbonate ion<Carbon dioxide<Carbon monoxide</u>
Answer:
work out if it's either going to sink or float
Explanation:
this can be carried out by calculating the numbers
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
Picometre is equal to 1·10⁻¹²m
175·10⁻¹² m = 1,75·10⁻¹⁰ m
