Answer:
5.916x10⁻³ mol OH⁻
Explanation:
The reaction that takes place is:
- H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
First we <u>calculate the added moles of each reagent</u>, using the <em>given volumes and concentrations</em>:
- H₂SO₄ ⇒ 0.144 M * 27.55 mL = 3.967 mmol H₂SO₄
- KOH ⇒ 0.316 M * 43.84 mL = 13.85 mmol KOH
Now we<u> calculate how many KOH moles reacted with 3.967 mmol H₂SO₄</u>:
- 3.967 mmol H₂SO₄ *
= 7.934 mmol KOH
Finally we calculate how many OH⁻ moles remained after the reaction
- 13.85 mmol - 7.934 mmol = 5.916 mmol OH⁻
- 5.916 mmol / 1000 = 5.916x10⁻³ mol OH⁻
Answer:
F
O
N
Mg
Explanation
The key explanation here is that we need to know the trends in electronegativity value increase or decrease across the periods and down the group.
Across the periods, electronegativity is expected to increase. Hence, with the exception of the noble gases, the most electronegative elements are found further right on the periodic table. While electronegativity is expected to increase across the period, it decreases down the group.
Hence, if two elements are in the same group, further down we have the one with the less electronegative value. If two elements are in the same period, further right we have the more electronegative element.
If it's an endothermic reaction, then that means heat is being added to the system therefore H>0. Entropy is disorder, and since there are more moles on the products side, entropy is increasing therefore S>0 as well.