B) both poles
Plz Put me as brainiest
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:
Reduction
Explanation:
Gaining electrons means reduction. When the electrons are on the reactant side of the reaction, they are added to the other reactant. This indicates that the Sn^4+ is gaining electrons and is therefore reduced (reduction).
Answer: 41.46 L
Explanation:
La ecuación que describe relación entre presión, volumen, temperatura y la cantidad (en moles)
de un gas ideal es:
PV = nRT
Donde: P = Presión absoluta
, V= Volumen , n = Moles de gas
, R = Constante universal de los gases ideales, T = Temperatura absoluta,
R = 0.082 L. atm/mol. °K
V = nRT/P
Calculanting n
n = mass/ molecular mass
<h3>n = 4 g / 2g. mol⁻¹</h3><h3>n = 2 mol</h3><h3>T =25⁰ + 273 ⁰K = 298 ⁰K</h3><h3>V = (2 mol ₓ0.082 L. atm / mol.°K x 298 ⁰K) / 1.18 atm = 41.46 L</h3>
Answer:
The answer would be It breaks them up.
Explanation: