Explanation:
Compounds having same molecular formula but different structural and spatial arrangement are isomers.
Three isomers are possible for dibromomethene.
In one structure (IUPAC name: 1,1-dibromomethene), both the bromine atoms are attached to one carbon atom.
In another two structures (Cis and trans), two bromine atoms are attached to two different carbon atoms.
In Cis 1,2-dibromomethene, two bromine atoms are present on the same side.
Whereas in Cis 1,2-dibromomethene, two bromine atoms are present on the opposite side and hence, does not have net dipole moment.
The formula we can use here is the Plancks equation:
E = h c / ʎ
where h is Plancks constant = 6.626 × 10-34 m2 kg / s, c
is speed of light = 3 x 10^8 m/s and ʎ is wavelength = 656.1 x 10^-9 m
Therefore E is:
E = (6.626 × 10-34 m2 kg / s)
* (3 x 10^8 m/s) / 656.1 x 10^-9 m
<span>E = 3.03 x 10^-19 J</span>
Answer:
20 m/s^2
Explanation:
given,
final velocity (v) = 6000m/s
initial velocity (u) = 0m/s
time taken (t) = 5 minutes
= 5×60second
= 300second
acceleration(a) = ?
we know that,
a = (v-u)/t
= (6000-0)/300
= 20 m/s^2
Answer:
The type of reaction for the following equation is combustion equation.
Explanation:
Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.
The reaction given to us:
When 1 mole of ethane reacts with 7/2 moles of oxygen gas it gives 3 moles of water and 2 moles of carbon dioxide gas.
The type of reaction for the following equation is combustion equation.
Problem 2
You start out with 216 ugrams of Fermium - 253. After 3 days, you will have 1/2 as much. 108 ugrams is what you have.
Another 3 days goes by. You started with 108 ugrams. That gets cut in 1/2 again. Now you have 54 ugrams.
Finally another 3 days goes by. You started with 54 ugrams. you now have 1/2 as much which would be 27 ugrams
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
You are using Nitrogen as your base example. The first thing you should do is fill in the table. Then you should try and make some rules. You need the rules in case the exam you are preparing for picks a different element to talk about these bond tendencies. In any event, it's handy to think this way.
<em><u>Table</u></em>
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
<em><u>Rules</u></em>
As the number of bonds INCREASES, the energy contained in the bond goes UP
As the number of bonds INCREASES, the length of the bond goes DOWN.
