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3 years ago

Which two things affect how a lens refracts light?

Chemistry

2 answers:

gayaneshka [121]3 years ago

7 0

Answer:

hey hope this work

The amount of bending depends on two things: Change in speed – if a substance causes the light to speed up or slow down more, it will refract (bend) more. Angle of the incident ray – if the light is entering the substance at a greater angle, the amount of refraction will also be more noticeable. 

 and the answer i think is A

Romashka-Z-Leto [24]3 years ago

5 0

Answer:

Explanation:

the correct answer is A

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Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0

3 years ago

4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o

Vilka [71]

Answer: The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol$

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

$\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}$

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = $\frac{1}{1}\times 0.0711=0.0711mol$ of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

$0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g$

To calculate the percentage yield of 3-hexanol, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

$\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%$

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

4 0

3 years ago

V = d/t

dangina [55]

Answer:

Explanation:

Did you mean: V = d/t a = (V - Vit Average = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?

Showing results for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = "-9.81" m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?

Search instead for V = d/t a = (V - Vil/t Vaverage = (V+ + V)/2 with constant acceleration d = Vit + 2 at? Vi = (V2 + 2ad)1/2 =VV2 + 2ad A stick figure throws a ball straight up into the air at 5 m/s. g = -9.81 m/s2 1. How long does it take to reach the top? 2. How long does it take to come back to the level of release? 3. If the hand is 1 m from the ground, how long will it take to hit the ground if the ball is not caught? 4. How high is the ball at the top from the ground? 5. What is the displacement of the ball, if it is caught on return? 6. What is the displacement of the ball to the top from release? 7. What is final velocity when you catch the ball on return to your hand? 8. What is the final velocity as it hits the ground? 9. What is the velocity at the top?

7 0

3 years ago

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AnnyKZ [126]

Answer:

Already answered brainly.com/question/845321563

Explanation:

8 0

3 years ago

What is the main function of the part labeled Y in the model?

Tanya [424]

Answer