The study strategy Lauren is using in spreading her study sessions over a period of time is pacing, which helps the student develop a schedule focused on their own study pace.
<h3>Pacing Study Sessions</h3>
This study strategy of distributing the study into short sessions rather than studying the entire content through one long session is more effective in retaining content and learning.
What happens is that Lauren is using mass repetition processing, which can be compared to a longitudinal wave in physics, with spaces in between, concentrating the initial review close to the proof to ensure retention and avoid forgetting.
Through pacing, Lauren achieves greater motivation to carry out her studies in a concentrated and focused way, helping her to retain and preserve knowledge.
The correct answer is:
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Answer:
6,8 g
Explanation:
c = 4.18 J/(g * °C) = 4180 J / (kg * °C)
= 25 °C
= 36,4 °C
Q = 325 J
The formula is: Q = c * m * (
)
m = 
Calculating:
m = 325 / 4180 * (36,4 - 25) ≈ 0,0068 kg = 6,8 g
Answer:
2Au₂S₃ + 6H₂ → 4Au + 6H₂S
Explanation:
Balancing:
2Au₂S₃ + 6H₂ → 4Au + 6H₂S
Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.
For the compound, 
the electronic configuration of the atoms, carbon and hydrogen are:
Carbon (atomic number=6): In ground state= 
In excited state: 
Hydrogen (atomic number=1): 
All the bonds in the compound is single bond(
-bond) that is they are formed by head on collision of the orbitals.
The structure of the compound is shown in the image.
The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.
In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in of hydrogen will overlap to the 2p^{3}-orbital of carbon.
Thus, the hybridization of Hydrogen is 
-hybridization and the hybridization of Carbon is 
-hybridization.
The hybridization of each atom is shown in the image.
The half-life equation is written as:
An = Aoe^-kt
We use this equation for the solution. We do as follows:
5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE
Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?
G(t) = 176e^-0.02t
How many grams of goo will remain after 50 minutes?
G(t) = 176e^-0.02(50) = 64.75 g
