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Ugo [173]
4 years ago
10

Which of the following terms describes the motion of particles during the transmission of wave energy? A. non-recurrent B. arbit

rary C. periodic D. Unpredictable
Chemistry
2 answers:
Vlad [161]4 years ago
5 0
The term that describes the motion of particles during the transmission of wave energy would be C. periodic.
Hope this helps!
oee [108]4 years ago
5 0

Answer:

C. periodic.

Explanation:

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How did the peak of Mt. Everest rise from sea level to 8,848 feet?
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Answer:

The height of Everest. Controversy over the exact elevation of the summit developed because of variations in snow level, gravity deviation, and light refraction. The figure 29,028 feet (8,848 metres), plus or minus a fraction, was established by the Survey of India between 1952 and 1954 and became widely accepted.

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3 years ago
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What is the final [Na+] in a solution prepared by mixing 70.0 mL of 3.00 M Na2SO4 with 30.0 mL of 1.00 M NaCl?
Lynna [10]

Answer:

4.5 M

Explanation:

70.0 ml was mixed in 3.00 M of Na2SO4

30.0 ml was mixed in 1.00 M of NACL

The first step is to convert 70 ml to liters

= 70/1000

= 0.07 liters

The formular for molarity is

moles/liters

The number of moles in Na2S04 can be calculated as follows

Let y represent the number of moles

3M= y moles/0.07

= 3×0.07

= 0.21 moles

Since Na2So4 has 2 moles of Na then the number of moles is

= 2×0.21

= 0.42 moles

Convert 30ml to liters

= 30/1000

= 0.03 liters

The number of moles in Nacl can be calculated as follows

Let y represent the number of moles

1M= y moles/0.03

= 1×0.03

= 0.03 moles

Since Nacl has 1 mole of Na then the number of moles is

= 1 × 0.03

= 0.03 moles

Therefore the final Na+ can be calculated as follows

Total moles = 0.03 moles + 0.42 moles

= 0.45 moles

Total liters= 0.07 liters + 0.03 liters

= 0.1 liters

Na+ = 0.45/0.1

= 4.5 M

Hence the final Na+ in the solution is 4.5 M

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A liquid takes 10.14 x 10^6 J of energy to boil 28.47 kg at 298 K. Using the latest heats of vaporization, what substance is thi
otez555 [7]

A liquid takes 10.14 x 10^6 J of energy to boil 28.47 kg at 298 K. Using the Latent heats of vaporization of some of the substances:

Acetone: 538,900 \frac{J}{kg}

Ammonia: 1,371,000 \frac{J}{kg}

Propane: 356,000 \frac{J}{kg}

Methane: 480,600 \frac{J}{kg}

Ethanol: 841,000 \frac{J}{kg}

Calculating the latent heat of vaporization of the given substance:

Heat required = 10.14 * 10^{6} J

Mass of the substance boiled = 28.47 kg

Latent heat of vaporization =\frac{10.14 * 10^{6} J}{28.47 kg}  = 3,56,000 \frac{J}{kg}

So the substance boiled is propane.


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