Answer:
150 g/mol
Explanation:
Let's consider the complete neutralization of a diprotic acid H₂X with NaOH.
H₂X + 2 NaOH → Na₂X + 2 H₂O
40.0 mL of 0.200 M NaOH. were required to reach the endpoint. The reacting moles of NaOH are:
0.0400 L × 0.200 mol/L = 8.00 × 10⁻³ mol
The molar ratio of H₂X to NaOH is 1:2. The reacting moles of H₂X are 1/2 × 8.00 × 10⁻³ mol = 4.00 × 10⁻³ mol.
4.00 × 10⁻³ moles of H₂X have a mass of 0.600 g. The molar mass of H₂X is:
0.600 g/4.00 × 10⁻³ mol = 150 g/mol
Answer:
1.30 g
Explanation:
La reacción que toma lugar es:
Primero <u>convertimos 3.28 g de MgO en moles</u>, usando su <em>masa molar</em>:
- 3.28 g ÷ 40.3 g/mol = 0.081 mol MgO
Después <u>convertimos 0.081 moles de MgO en moles de O₂</u>, usando los <em>coeficientes estequiométricos</em>:
- 0.081 mol MgO *
= 0.0407 mol O₂
Finalmente<u> convertimos 0.0407 moles de O₂ en gramos</u>:
- 0.0407 mol O₂ * 32 g/mol = 1.30 g
-A bromine with an incomplete octet and a positive charge.
-A bromine with an incomplete octet and a positive charge and an oxygen with ten valence electrons and two charges.
-A compound with the bond between a bromine and an oxygen with three bonds.
