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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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When an ionic solid is added into a solvent, you can see that the ionic solid dissociates into its respective cations and anions
joja [24]

Answer: See explanation

Explanation:

Based on the information given in the question, the ranking from the first step to the last step goes as follows:

1. Dissolution of salt into cations and anions

2. Hydration of anions

3. Hydration of cations

4. Dissolved cations and anions begin to deposit as a solid salt.

5. Rate of dissolution is equal to the rate of recrystallization.

4 0
3 years ago
During an experiment, Mr. Lehman tells his students to be alert for signs that an endothermic reaction is happening. Which of th
Aliun [14]

A test tube becomes cool to the touch.

Explanation:

The signs for an endothermic reaction would most likely be that a test tube becomes cool to touch. An endothermic reaction is a reaction in which heat is absorbed from the environment.

  • At the end of this heat change, the surrounding becomes colder.
  • Most endothermic reactions are always shown by the test tube becoming colder.
  • For exothermic reaction in which heat is given off to the surroundings, the tube should be hotter.

Learn more:

Exothermic reactions brainly.com/question/10567109

#learnwithBrainly

4 0
3 years ago
Read 2 more answers
Using the bond energy data from your text (or the internet), determine (show calculations for) the approximate enthalpy change ,
aleksley [76]

Answer:

∆H=  <u>438 KJ/mol</u>

Explanation:

First, we have to find the <u>energy bond values</u> for each compound:

-) Cl-Cl = 243 KJ/mol

-) F-F = 159 KJ/mol

-) F-Cl = 193 KJ/mol

If we check the reaction we can calculate the <u>number of bonds</u>:

Cl_2_(_g_)~+~3F_2_(_g_)~->~2ClF_3_(_g_)

In total we will have:

-) Cl-Cl = 1

-) F-F = 3

-) F-Cl = 6

With this in mind. we can calculate the <u>total energy for each bond</u>:

-) Cl-Cl = (1*243 KJ/mol) = 243 KJ/mol

-) F-F = (3*159 KJ/mol) = 477 KJ/mol

-) F-Cl = (6*193 KJ/mol) = 1158 KJ/mol

Now, we can calculate the total energy of the <u>products</u> and the <u>reagents</u>:

Reagents = 243 KJ/mol + 477 KJ/mol = 720 KJ/mol

Products = 1158 KJ/mol

Finally, to calculate the total enthalpy change we have to do a <u>subtraction</u> between products and reagents:

∆H= 1158 KJ/mol-720 KJ/mol = <u>438 KJ/mol</u>

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I hope it helps!

7 0
3 years ago
A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glo
allsm [11]

Answer:

PART A: 412.98 nm

PART B: 524.92 nm

Explanation:

The equation below can be used for a diffraction grating of nth order image:

n*λ = d*sinθ_{n}

Therefore, for first order images, n = 1 and:

λ = d*sinθ_{1}.

The angle θ_{1} can be calculated as follow:

tan θ_{1}  = 9.95 cm/15.0 cm = 0.663 and

θ_{1} = tan^{-1} (0.663) = 33.56°

Thus: d =λ/sin θ_{1}  = 461/sin 33.56° = 833.97 nm

PART A:

For a position of 8.55 cm:  

tan θ_{1}  = 8.55 cm/15.0 cm = 0.57 and

θ_{1} = tan^{-1} (0.57) = 29.68°

Therefore:

λ =d*sin θ_{1}  = 833.97*sin 29.68° = 412.98 nm

PART B:

For a position of 12.15 cm:  

tan θ_{1}  = 12.15 cm/15.0 cm = 0.81 and

θ_{1} = tan^{-1} (0.81) = 39.01°

Therefore:

λ =d*sin θ_{1}  = 833.97*sin 39.01° = 524.92 nm

6 0
3 years ago
Please someone help meeeeee!!!
Lorico [155]

Answer:

1 = Q =  7315 j

2 =Q =  -21937.5 j

Explanation:

Given data:

Mass of water = 50 g

Initial temperature = 20°C

Final temperature = 55°C

Energy required to change the temperature = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 j/g.°C.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 55°C - 20°C

ΔT = 35°C

Q = 50 g× 4.18 j/g.°C×35°C

Q =  7315 j

Q 2:

Given data:

Mass of metal = 100 g

Initial temperature = 1000°C

Final temperature = 25°C

Energy released = ?

Specific heat capacity = 0.225 j/g.°C

Solution:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 25°C - 1000°C

ΔT = -975°C

Now we will put the values in formula.

Q = 100 g × 0.225 j/g.°C × -975°C

Q =  -21937.5 j

Negative sign show that energy is released.

7 0
3 years ago
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