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Marina CMI [18]
3 years ago
15

A student took notes in class. - Uses high frequency sound waves - Creates images from echoes - Has many applications - Medical

- Pregnancy - Kidney stones - Cancer - Injury identification - Other - Cleaning - Identification of objects - Breakdown of bacteria - Production Which would be the best title for these notes?
Echolocation
Ultrasound Technology
Digital Signal
Analog Signal
Chemistry
2 answers:
SpyIntel [72]3 years ago
5 0
The correct option should be ultrasound technology (option B) because it is related to sonographers or ultrasound technicians. they are most likely with while pregnancy but they have plenty of uses, such as evaluating and diagnosis, many medical treatment for elderly patients. 
Fudgin [204]3 years ago
3 0
Ultrasound technology 
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What mass of gold is produced when 23.8 a of current are passed through a gold solution for 36.0 min ?
SVETLANKA909090 [29]

Answer:-

Solution:- The equation for the conversion of gold ions to gold metal is as follows:

Au^3^+(aq)+3e^-\rightarrow Au(s)

From this equation, 1 mol of Gold metal is deposited by 3 mole of electrons.

We know that one mol of electron carries one Faraday that is 96500 Coulombs.

So. 3 moles of electrons would be = 3(96500) = 289500 Coulombs

We could calculate the total Coulombs by using the formula:

q = i*t

where, q is the charge in coulombs, i is the current in ampere and t is time in seconds.

q = 23.8 ampere*36.0 min(\frac{60 sec}{1 min})

q = 51408 ampere per second

Since, 1 ampere per second = 1 coulomb

So, 51408 ampere per second is same as 51408 coulombs.

Let's use this value of q and the info from balanced equation and molar mass(196.967 g epr mol) of gold to calculate the mass of it being deposited while the current is passed as:

51408 coulomb(\frac{1 mol Au}{289500 coulombs})(\frac{196.967 g}{1 mol})

= 35.0 g Au

So, 35.0 g of gold are deposited on passing 23.8 ampere of current for 36.0 min to a gold solution.


7 0
3 years ago
In plants, this is period in which a seed or plant does not grow, waiting for the necessary environmental conditions like the ri
Lunna [17]

Answer:

The answer is Germination.

Explanation:

During the stage of germination, the plant needs the right temperature, day length, amount of water, and nutrients.

8 0
4 years ago
Calculate molecules in 1dm^3 of oxygen
Archy [21]

Number of molecules in 1 dm³ Oxygen = 2.71 x 10²²

<h3>Further explanation</h3>

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.

The mole is the number of particles contained in a substance

1 mol = 6.02.10²³

1 dm³ of oxygen = 1 L Oxygen

  • mol Oxygen :

\tt \dfrac{1}{22.4}=0.045`mol

  • molecules of Oxygen :

n=mol=0.045

No = 6.02.10²³

\tt N=n\times No\\\\N=0.045\times 6.02\times 10^{23}\\\\N=2.71\times 10^{22}

5 0
3 years ago
__________________ are high-speed bands of winds occurring at the top of the tropsophere.
chubhunter [2.5K]

Answer:

jet streams

Explanation:

THE JET STREAM Narrow bands of exceedingly high speed winds are known to exist in the higher levels of the atmosphere at altitudes ranging from 20,000 to 40,000 feet or more. They are known as jet streams.

7 0
3 years ago
Ср Substance Aluminum Calcium Water Potassium Zinc Cp 0.895 J/g^ C 0.650 J/g ° 4.184 J/g^ C 0.750 J/g^ C 0.388J / (g ^ 0) * C Su
BabaBlast [244]

Answer:

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Explanation:

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That's the question right

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