The reaction
→
is best classified as double displacement reaction.
Those reaction in which two compounds react by exchanges of ions to form two new compounds is called double displacement reaction. The easiest way to identify double displacement reactions is to check to see whether the cations exchanged anions with each other or not . Always balanced chemical equation is used to determine.
There are three types of double displacement reaction which is given as,
- Precipitation
- Neutralization
- Gas formation
The real world example of double displacement reaction is combining vinegar and baking soda to create homemade volcano.
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Answer:
because the electron is a negatively charge and have high energy
1) Carbon-13:
Proton-6 Neutron-7 Electron-6
2)Atomic mass of element X:
(55*10+56*20+57*70)/100=56.6
Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Answer:
4
Explanation:
Ionization energy can be defined as the energy required for an atom to lose its valence electron to form an ion. Hence, it deals with how easily an atom would lose its electron and form an ion. As the valence electrons are lossless bound to the outermost shell, they can easily be lost without much problem or better still they can be lost easily. Hence, the energy change here is small and thus we can conclude that the ionization energy here is low.
The electron affinity works quite differently from the ionization energy. It deals with the way in which a neutral atom attracts an electron to form an ion. For an electron with loose valence electrons, the sure fact is that it does not really need these electrons. Hence, there is no need for an high electron affinity on its part. Thus, we conclude that the electron affinity is also low