Answer: Belongs to the group 2A
Explanation:
As you can see, the first two ionization energies are close and low, meaning that this element ionizates easily.
Not only loses easily the first electron, but the second too
To remove the third electron you requiered a huge amount of energy
Now, elements easily ionizable are the ones from group IA, group 2A and transition metals.
The last ones have mixed characteristics in matter of how many electrons you can remove from them, so they are not a family.
Now the question: group I or group II ?
The elements of group I have low ionization energies for the first electron but high energies for the second ones.
Being all that said, the unknown element belongs to the Group 2A
Correct answer to this question is c. water evaporating
Explanation:
The rate of reaction is the speed at which a chemical reaction takes place, defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time. Reaction rates can vary dramatically.
The missing part of the equation is found to be 4/2He. Option A
<h3>What are nuclear equations?</h3>
The term nuclear equations have to do with the type of equation in which one type of nucleus is transformed into another sometimes by the bombardment or loss of a particle.
Now the full equation ought to be written as 7/3Li + 1/1H -----> 4/2He + 4/2He. This is because the total mass on the left is 8 and the total charge on the left is 4.
Learn more about nuclear equations:brainly.com/question/19752321
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Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
