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3 years ago

14

How is a wave propagated and produced

1 answer:

juin [17]3 years ago

4 0

Answer:

A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

Explanation:

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3. The opening that allows light to enter the eye is the

xxMikexx [17]

Answer:

A) Iris

Explanation:

The Iris controls the diameter of the opening in your eye, thus also controlling the amount of light let inside. It's also reponsible for the colour of you eye.

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3 years ago

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How many moles of water can be produced by reacting 525 liters of oxygen with excess hydrogen

Ann [662]

I believe the answer is 16

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3 years ago

Calculate the number of moles in 25.0 g of each of the

mamaluj [8]

Answer:

A. 6.25moles

B. 0.78moles

C. 0.32moles

D. 0.15moles

E. 0.43moles

Explanation:

use

n = mass/molar mass

A. He

25/4 = 6.25moles

B. O2

25/32 = 0.78moles

C. Al(OH)3

27+ 3(17) = 78
25/78 = 0.32 moles

D. GaS3

70+3(32) = 166
25/166 = 0.15moles

E. C4H10

4(12)+10(1) =58
25/58 = 0.43moles

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3 years ago

Which best describes two ways in which hydrogen is involved in photosynthesis?

sveticcg [70]

Answer:

C

Explanation:

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3 years ago

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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