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nevsk [136]
3 years ago
14

To locate objects in their environments, bats in flight and porpoises under water both use ultrasound waves with frequencies tha

t are beyond human hearing. These animals produce an ultrasonic wave and then detect echoes from nearby objects in order to determine the location of the object Porpoise Bat If a porpoise and a bat both produce ultrasonic waves when they are 16 meters from an object, which animal would hear its echo first? The bat would hear its echo first because sound travels faster in air than in water. O The porpoise would hear its echo first because sound travels faster in water than in air. O The bat would hear its echo first because the frequency of sound waves is greater in air than in water 0 The porpoise would hear its echo first because the frequency of sound waves is greater in water than in air.​
Chemistry
1 answer:
Katarina [22]3 years ago
3 0

Answer: The porpoise would hear its echo first because sound travels faster in the water than in air.

Explanation:

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What is discriminate?
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Answer:

Explanation:

To make a distinction in favor of or against a person or thing on basic or group, class or category, to which the person or thing(s) belong

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3 years ago
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3 years ago
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15. Within a voltaic cell,___
kobusy [5.1K]

Answer:

B. oxidation; reduction

Explanation:

A voltaic cell is electro-chemical cell in which chemical energy is converted into electrical energy.

1. This cell utilizes chemical reaction to generate electric.\

2. there two electrode anode and cathode

3. At Anode oxidation occurs

4. At cathode reduction occurs

5. chemical is present in the cell which is electrolyte which completes the circuit of the voltaic cell.

  • oxidation is the process in which there is loss of electrons
  • Reduction is the process in which there is gain of electrons

___________________________________________________

Based on above discussion

At anode oxidation takes place

At cathode reduction takes place.

Hence, correct option is B. oxidation; reduction

7 0
3 years ago
A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi
Aleksandr-060686 [28]

Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K

n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>

<u><em /></u>

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = <em>9.9652g of water</em>

<em />

I hope it helps!

4 0
3 years ago
A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 300°C. How many moles of argon gas does the cylinder co
KengaRu [80]
In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.

Given 
P = 10130.0 kPa 
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol

Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.

PV = nRT                      PV
----    ------    --->    n = --------
 RT     RT                       RT

          10130.0 kPa  x 50 L
n= ---------------------------------------------
       8.314 L. kPa/K.mol  x 573.15 K
             506,500 
  =  ----------------------------
         4,765.17  mol K

=106.29 mol Ar

So the moles of argon gas is 106.29 moles 
8 0
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