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natita [175]
3 years ago
6

A key difference between the James-Lange theory and the Cannon-Bard theory of emotion is that in the first, physiological change

s _____; in the latter, physiological changes _____.
Chemistry
1 answer:
mariarad [96]3 years ago
7 0

Answer:

<h2>1. precede emotional experience</h2><h2>2. occur at the same time</h2>

Explanation:

William James and Carl Lange gave a theory about a type of emotion is called a James-Lange theory. According to this theory, physiological arousal creates an emotional experience. According to this theory, physiological changes are the primary condition while emotion is felt when brain interacts with the information obtained from the body.  

                    According to the Cannon-Bard theory, the emotional feeling occurs when the dorsal thalamus is stimulated. In this theory, physiological changes and emotional experiences take place at the same time.      

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Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
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Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

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\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

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