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:
Percent Yield Fe = 82.5%
Explanation:
The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.
To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.
Atomic Mass (Mg): 24.305 g/mol
Atomic Mass (Fe): 55.845 g/mol
3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂
20.5 g Mg 1 mole 2 moles Fe 55.845 g
----------------- x ----------------- x ---------------------- x ----------------- =
24.305 g 3 moles Mg 1 mole
= 31.4 g Fe
Actual Yield
Percent Yield = ---------------------------------- x 100%
Theoretical Yield
25.9 g Fe
Percent Yield = -------------------- x 100%
31.4 g Fe
Percent Yield = 82.5%
Answer:
(A) 0.129 M
(B) 0.237 M
Explanation:
(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:
- 2HA + Ba(OH)₂ → BaA₂ + 2H₂O
Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).
We <u>convert mass of phthalate to moles</u>, using its molar mass:
- 0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol
Now we <u>convert mmol of HA to mmol of Ba(OH)₂</u>:
- 9.27 mmol HA *
= 6.64 mmol Ba(OH)₂
Finally we calculate the molarity of the Ba(OH)₂ solution:
- 6.64 mmol / 35.8 mL = 0.129 M
(B) The reaction between Ba(OH)₂ and HCl is:
- 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O
So<u> the moles of HCl that reacted </u>are:
- 17.1 mL * 0.129 M *
= 4.41 mmol HCl
And the <u>molarity of the HCl solution is</u>:
- 4.41 mmol / 18.6 mL = 0.237 M
