Increases
Stays the same
s-orbitals
d-orbitals
p-orbitals
Explanation:
The trend in the electronic configuration on the periodic table follows a definite pattern because the arrangement of elements on the table is based on the periodic law.
The periodic law states that "The properties of elements are a function of their atomic number".
- Elements in the same horizontal period have increasing number of valence electrons.
- The number of valence electrons stays the same down a group.
- Groups 1 and 2 are known as the s-block and electrons are added to the s-orbitals for them.
- Groups 3 to 12 is the transition block and electrons the valence d-orbital.
- Groups 3 through 18 is the p-block where the valence shell is the p-orbital.
learn more:
Periodic table brainly.com/question/8543126
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Heat = 1.74 kJ
<h3>Further explanation</h3>
Given
melts at 328 ℃ + 273 = 601 K
mass = 23 g = 0.023 kg
initial temperature = 297 K
Final tmperature = 702 K
Required
Heat
Solution
1. raise the temperature(297 to 601 K)
c of lead = 0.130 kJ/kg K
Q = 0.023 x 0.13 x (601-297)
Q = 0.909 kJ
2. phase change(solid to liquid)
Q = m.Lf (melting/freezing)
Q = 0.023 x 23 kj/kg = 0.529 kJ
3. raise the temperature(601 to 702 K)
Q = 0.023 x 0.13 x (702-601)
Q = 0.302 kJ
Total heat = 1.74 kJ
Im going with c hope this helps
Answer:
Explanation:
Equation for the heterogeneous system is given as:
⇄ 
The concentrations and pressures at equilibrium are:
![[A] = 9.68*10^{-2}M](https://tex.z-dn.net/?f=%5BA%5D%20%3D%209.68%2A10%5E%7B-2%7DM)
![[C]=14.64M](https://tex.z-dn.net/?f=%5BC%5D%3D14.64M)
![[D]=10.11M](https://tex.z-dn.net/?f=%5BD%5D%3D10.11M)
If we convert both pressure into bar; we have:
1 torr = 0.001333 bar
![K=\frac{[P_E]^3}{[A]^2[P_B]^3}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BP_E%5D%5E3%7D%7B%5BA%5D%5E2%5BP_B%5D%5E3%7D)
Explanation:
a. CGS system: centimetre as the unit of length, the gram as the unit of mass, and the second as the unit of time
b. SI system: metre as the unit of length, the kilogram as the unit of mass, and the second as the unit of time
