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Morgarella [4.7K]
3 years ago
8

When a horn is moving away from an observer the sound waves from it will pass by the observer____ frequently, causing the pitch

to____
Chemistry
1 answer:
xxMikexx [17]3 years ago
8 0
The answer: less frequency, causing the pitch to be lowered. 

This is known as the Doppler effect. When you have an object that is producing sound while moving, the side that it is moving towards to will have waves that will become a bit more squished together, while the other side that it is moving away from will have waves that are spaced a bit more apart from each other. The only determining factor in this is the magnitude of the speed, so the faster the object is moving towards or away from you, then the more severe pitch change the object will have.
In this case, it's just asking in general what would happen if the horn was moving away from you while it was giving off sound waves. Therefore, since it is moving away from you, its sound waves are a bit further apart, resulting in a lowered frequency and pitch.
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Explain the general process of nuclear fission. What is created from fission?
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2 years ago
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How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?
ankoles [38]
The chemical reaction equation for this is 

XeF6 + 3H2 ---> Xe + 6HF

Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.

we get n = 0.1433 moles H2

to get the mass of XeF6, 

we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.

0.1433 moles H2 x \frac{1 mole XeF6}{3 moles H2} x \frac{245.28 g XeF6}{1 mole XeF6} = 11.71 g XeF6

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3 years ago
For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.
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Answer:

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The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

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The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

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