The Step that is Usually done right before the experimentation in the Scientific Method is Developing a Hypothesis and asking a Question.
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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
Answer:
10.87 g of Ethyl Butyrate
Solution:
The Balance Chemical Equation is as follow,
H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH → H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O
According to equation,
88.11 g (1 mol) Butanoic Acid produces = 116.16 g (1 mol) Ethyl Butyrate
So,
8.25 g Butanoic Acid will produce = X g of Ethyl Butyrate
Solving for X,
X = (8.25 g × 116.16 g) ÷ 88.11 g
X = 10.87 g of Ethyl Butyrate
Get the molarity we need to divide the number of moles of NaCl by the volume of the solution. So, 0.32 moles NaCl divided by 3.4 L, and that gives 0.094 M NaCl.
Answer:
Element Atomic Number One number you will find on all periodic tables is the atomic number for each element.
Explanation:
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