Answer:
Cuando dos átomos se acercan entre sí, se genera un compuesto al compartir pares de electrones que cada uno de los átomos aporta, permitiéndoles alcanzar los 8 electrones de valencia (octeto) en su capa externa.
Explanation:
La configuración electrónica del elemento puede escribirse de la siguiente manera;
1s²2s²2p⁴
La configuración electrónica dada es equivalente a la del oxígeno, por lo tanto, tenemos;
El número de electrones en la capa de valencia = 2 + 4 = 6 electrones
Por consiguiente, cada átomo necesita 2 electrones para completar sus 8 electrones (octeto) en la capa externa.
Al acercarse los dos átomos, reaccionan y se combinan para formar un compuesto al compartir 4 electrones, 2 de cada átomo, de modo que cada átomo obtenga 2 electrones adicionales en su órbita externa en el nuevo compuesto y así se logre la configuración estable de octeto para cada uno de los átomos en el compuesto recién formado.
Answer:
Both CaCl2 and CaBr2 consist of elements (bromine and chlorine) from the same group (group 7).
Explanation:
In the periodic table, elements are arranged into groups based on their valence electron count in the outermost shell. Elements in the same group, which possess a similar number of valence electrons, typically exhibit similar chemical behaviors.
Chlorine and Bromine in CaCl2 and CaBr2 belong to group 7, known as HALOGENS, characterized by having 7 valence electrons in their outer shell.
The similarity in properties between CaCl2 and CaBr2 arises because both contain Chlorine and Bromine, leading to analogous reactions and behaviors when interacting with other compounds.
Answer:
25.2 kJ
Explanation:
The full question can be found in the image linked to this response.
It's important to highlight that the heat absorbed by the 2.00 L of water for increasing its temperature from the beginning to the end comes solely from the burning of benzoic acid, as there are no heat transfers to the container or the surroundings.
To find the heat released from benzoic acid combustion, we simply measure the heat needed to warm the water.
Q = mCΔT
To find the mass of the water,
Density = (mass)/(volume)
Mass = Density × volume
Density = 1 g/mL
Volume = 2.00 L = 2000 mL
Mass = 1 × 2000 = 2000 g
C = specific heat of water = 4.2 J/g.°C
ΔT = (final temperature) - (Initial temperature)
<pAccording to the graph,
Final water temperature = 25°C
Initial water temperature = 22°C
ΔT = 25 - 22 = 3°C
Q = (2000×4.2×3) = 25,200 J = 25.2 kJ
Hope this Helps!!!
Problem 2
You begin with 216 micrograms of Fermium - 253. After three days, the quantity halves, resulting in 108 micrograms left.
Another three days pass. Beginning with 108 micrograms, this amount gets halved again, leaving 54 micrograms.
Finally, after another three-day span, starting from 54 micrograms, you again halve this amount to reach 27 micrograms.
#days Amount in micrograms
0 216
3 108
6 54
9 27
Problem One
Your example is Nitrogen. Begin by completing the table, then formulate some rules to help prepare for possible alternate elements in the test. This approach is quite useful.
Table
Bond Energy Kj/Mol Bond Length pico meters
N - N 167 145
N=N 418 125
N≡N 942 110
Rules
As the number of bonds INCREASES, the energy within the bond also INCREASES
As the number of bonds INCREASES, the distance of the bond DECREASES.
Answer:
The integer value of x in the hydrate is 10.
Explanation:

Molar concentration of the solution = 0.0366 M
Volume of the solution = 5.00 L
Moles of hydrated sodium carbonate = n


Weight of hydrated sodium carbonate = n = 52.2 g
Molar mass of hydrated sodium carbonate = 106 g/mol + x * 18 g/mol



By solving for x, we arrive at:
x = 9.95, approximating to 10
The integer x in the hydrate equals 10.