Answer: 169ml
Explanation
At constant temperature, PV is constant
Standard pressure is 101 kpa
150ml X 114kpa=?mlX101 kpa
?=114/101 X150 = 169 ml
Hydrogen bonding occurs when hydrogen is bonded to an oxygen or nitrogen or fluorine atom. In this case, the hydrogen atom in a hydrogen fluoride molecule will be able to bond to the fluoride atom of another hydrogen fluoride molecule, forming a hydrogen bond.
Answer:
mass (g) needed = 710.2 grams Na₂SO₄(s)
Explanation:
Needed is 2.5 Liters of 2.0M Na₂SO₄; formula wt Na₂SO₄ = 142.04g/mol.
mass (grams) of Na₂SO₄(s) = Molarity needed x Volume needed in Liters x Formula Wt of solute
mass (grams) of Na₂SO₄(s) = (2.5L)(2.0M)(142.04g/mol) = 710.2 grams Na₂SO₄(s)
Mixing: Transfer 710.4 grams Na₂SO₄ into mixing vessel and add water-solvent up to but not to exceed 2.5 Liters total volume. Mix until dissolved.
Gives 2.5 Liters of 2.0M Na₂SO₄(aq) solution.
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>
<h3><u>Answer</u>;</h3>
17.25 moles of hydrogen
<h3><u>Explanation;</u></h3>
The equation for the reaction;
2HI (g) → H₂(g) + i₂ (g)
34.5 moles of Hydrogen iodide are decomposed;
The mole ratio of HI to Hydrogen = 2 : 1
Therefore;
moles of hydrogen = (34.5/2) ×1
= 17.25 moles
Therefore; 17.25 moles of hydrogen will be produced when 34.5 moles of Hydrogen iodide are decomposed