(c) Cu + S → CuS is classified as a redox reaction
Explanation:
The following reactions are presented:
(a) K₂CrO₄ + BaCl₂ → BaCrO₄ + 2 KCl
(b) Pb²⁺ + 2 Br⁻ → PbBr₂
(c) Cu + S → CuS
Reaction (c) represents a redox reaction, as the oxidation states of the elements are changing. In this case:
Cu + S → CuS
In its elemental form, Cu has an oxidation state of 0, while in CuS (copper sulfide), its oxidation state changes to +2.
Similarly, S in its elemental form has an oxidation state of 0 and is -2 in CuS (copper sulfide).
Learn more about:
redox reactions
Respuesta:
0.16 M
Explicación:
Teniendo en cuenta:
O sea,
Dado que:
Para 
:
Molaridad = 0.200 M
Volumen = 20.0 mL
Convierte mL a L:
1 mL = 10⁻³ L
Entonces, volumen = 20.0×10⁻³ L
Los moles de son:
Moles de = 0.004 moles
Para 
:
Molaridad = 0.400 M
Volumen = 30.0 mL
Convertimos mL a L:
1 mL = 10⁻³ L
Volumen = 30.0×10⁻³ L
Entonces, los moles de son:
Moles de = 0.012 moles
Según la reacción:
1 mol de reacciona con 1 mol de
Por lo tanto,
0.012 mol de reacciona con 0.012 mol de
Moles disponibles de = 0.004 mol
El reactivo limitante es el que está en menor cantidad, entonces es el limitante (0.004 < 0.012).
La formación del producto depende del reactivo limitante, así que,
1 mol de reacciona con 1 mol de y produce 1 mol de 
0.004 mol de reacciona con 0.004 mol de y genera 0.004 mol de
Los moles restantes de son: 0.012 - 0.004 = 0.008 mol
El volumen total es 20 + 30 mL = 50 mL = 0.050 L
Por lo que la concentración del ion bario, 
, después de la reacción es:
Response:
H₂SO₄
Clarification:
Given a compound consisting of 0.475 g H, 7.557 g S, and 15.107 g O, we must compute the empirical formula by following specific steps.
Step 1: Compute the total mass of the compound
Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g
Total mass = 23.139 g
Step 2: Determine the percentage composition.
H: (0.475g/23.139g) × 100% = 2.05%
S: (7.557g/23.139g) × 100% = 32.66%
O: (15.107g/23.139g) × 100% = 65.29%
Step 3: Divide each percentage by the element's atomic mass
H: 2.05/1.01 = 2.03
S: 32.66/32.07 = 1.018
O: 65.29/16.00 = 4.081
Step 4: Normalize all values by the smallest one
H: 2.03/1.018 ≈ 2
S: 1.018/1.018 = 1
O: 4.081/1.018 ≈ 4
Thus, the empirical formula for the compound is H₂SO₄.
