Answer:
Are memories stored in just one part of the brain, or are they stored in many different parts of the brain? Karl Lashley began exploring this problem, about 100 years ago, by making lesions in the brains of animals such as rats and monkeys. He was searching for evidence of the engram: the group of neurons that serve as the “physical representation of memory” (Josselyn, 2010). First, Lashley (1950) trained rats to find their way through a maze. Then, he used the tools available at the time—in this case a soldering iron—to create lesions in the rats’ brains, specifically in the cerebral cortex. He did this because he was trying to erase the engram, or the original memory trace that the rats had of the maze.
Lashley did not find evidence of the engram, and the rats were still able to find their way through the maze, regardless of the size or location of the lesion. Based on his creation of lesions and the animals’ reaction, he formulated the equipotentiality hypothesis: if part of one area of the brain involved in memory is damaged, another part of the same area can take over that memory function (Lashley, 1950). Although Lashley’s early work did not confirm the existence of the engram, modern psychologists are making progress locating it. Eric Kandel, for example, spent decades working on the synapse, the basic structure of the brain, and its role in controlling the flow of information through neural circuits needed to store memories (Mayford, Siegelbaum, & Kandel, 2012).
Many scientists believe that the entire brain is involved with memory. However, since Lashley’s research, other scientists have been able to look more closely at the brain and memory. They have argued that memory is located in specific parts of the brain, and specific neurons can be recognized for their involvement in forming memories. The main parts of the brain involved with memory are the amygdala, the hippocampus, the cerebellum, and the prefrontal cortex
The balanced chemical reaction is expressed as:
<span>C3H8 + 5 O2 -> 3 CO2 + 4 H2O
We are given the amount of water produced. We use this amount for the calculations. We do as follows:
16.3 mol H2O ( 1 mol C3H8 / 4 mol H2O ) = 4.075 mol C3H8 needed
Hope this answers the question. Have a nice day.</span>
Answer:
Explanation:
Given that:
The chemical equation for the reaction is:
Br2(g) ⇌ 2Br(g)
Initially 0.0345M 0.0416M
![Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016](https://tex.z-dn.net/?f=Q_C%20%3D%20%5Cdfrac%7B%5BBr%5D%5E2%7D%7B%5BBr_2%5D%7D%20%3D%20%5Cdfrac%7B%280.0416%29%5E2%7D%7B%280.0345%29%7D%3D%200.05016)
Thus, the given reaction will proceed in the backward direction
The I.C.E table is as follows:
Br2(g) ⇌ 2Br(g)
I 0.0345 0.0416
C +x -2x
E (0.0345+x) (0.0416 -2x)
![K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BBr%5D%5E2%7D%7B%5BBr_2%5D%7D%20%3D%20%5Cdfrac%7B%280.0416-2x%29%5E2%7D%7B%280.0345%2Bx%29%7D%20%3D%200.00584)
= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)
= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x
= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²
= 0.00152908 - 0.17184x + 4x²
Solving by using Quadratic formula
x = 0.03038 or 0.0126
For x = 0.03038
At equilibrium
[Br₂] = (0.0345 + 0.03038) = 0.06488 M
[Br] = (0.0416 -2(0.03038)) = - 0.01916 M
Since we have a negative value for [Br], we discard the value for x
For x = 0.0126
At equilibrium
[Br₂] = (0.0345 + 0.0126) = 0.0471 M
[Br] = (0.0416 -2(0.0126)) = 0.0164 M
Answer:
d. Hydrophobic molecules are attracted to each other.
Explanation:
The term “hydrophobic effect” is associated with the spontaneous tendency of macromolecules, such as proteins, to prefer a conformation in an aqueous medium, with hydrophobic groups facing the interior of the mac romolecule, favoring attractive intramolecular interactions, and hydrophilic groups exposed on the surface, for maximize interactions with water molecules in the medium. This is because the hydrophobic molecules are attracted to each other, allowing them to turn inward.
<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>
<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>
<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>
<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>
<span>73.0<span>Tf</span>=2561 °C</span>
<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>
