Answer:
The configurations are illustrated below.
Explanation:
Hydrogen possesses a single electron in its outer shell, carbon has 4, nitrogen has 5, and oxygen holds 6. To achieve an octet (or duet for hydrogen), they require 1, 4, 3, and 2 electrons respectively.
Therefore, each hydrogen atom will share one electron with carbon, while the remaining electron will be shared with nitrogen, maintaining 4 electrons available for sharing. Carbon can form two bonds with both oxygen atoms, expanding its octet; however, this renders it unstable, leading to the formation of resonance structures (redistribution of electrons), and charge formation. One of the oxygen atoms will share only one electron with nitrogen.
The two structures are depicted below.
(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
Answer:
The nichrome wire has contaminants.
The sample solution might be tainted.
Explanation:
If the nichrome wire is contaminated, sodium impurities could be causing the yellow flame. The wire is initially placed in the flame without the sample to check for such impurities.
The testing solution could also be contaminated, causing it to display a color different from the anticipated shade of the test ion.
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.
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:
