Response:
Examine the explanation
Explanation:
In this scenario, Nitrogen (N) experiences oxidation while Chlorine (Cl) undergoes reduction.
In response to your inquiry:
N is oxidized from an oxidation state of -3 to -1.
Cl is reduced from an oxidation state of +1 to -1.
At this point,
Borneol will likely exhibit a lower Rf due to its boiling point.
Response:
V1 = 20.3L
Clarification:
P2 = 811.4Pa
V2 = 25.6L
P1 = 1023.6Pa
V1 =?
To answer this query, we will utilize Boyle's law, which states that the volume of a gas at constant temperature is inversely related to its pressure.
In mathematical terms,
V = k / P, where k = PV
The relationship can be defined as P1 × V1 = P2 × V2 = P3 × V3 =......=Pn × Vn
This simplifies to P1 × V1 = P2 × V2
Let’s rearrange for V1
V1 = (P2 × V2) / P1
Substituting values gives
V1 = (811.4 × 25.6) / 1023.6Pa
So, V1 = 20771.84 / 1023.6
This results in V1 = 20.29L, rounded to 20.3L
1) The chemical equation is
Cu + 2AgNO3 ---> Cu (NO3)2 + 2Ag
2) Molar ratios are as follows:
1 mol Cu: 2 moles AgNO3: 1 mol Cu (NO3)2: 2 mol Ag
3) Converting 12.83 * 10^23 atoms of Cu to moles gives:
12.83 * 10^23 atoms / (6.02 * 10^23 atoms / mol) = 2.131 mol Cu
4) Using the ratios:
2.131 mol Cu * 2 mol Ag / 1 mol Cu = 4.262 mol Ag
5) To convert 4.262 mol of silver to grams, use the atomic weight of silver:
mass = moles × atomic mass = 4.262 mol * 107.9 g / mol = 459.9 grams
Answer: 459.9 g
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:
