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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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..........The answer is B
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3 years ago
Two students are working together to build two models. Both models will represent the molecular structure of sodium bicarbonate,
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Answer:

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

<em>Note: Since no specific color was stated for oxygen atoms, the answer assigns blue colored jellybeans to represent oxygen atoms.J</em>

Explanation:

Sodium bicarbonate, NaHCO₃ is a compound composed of one atom of sodium, one atom of hydrogen, one atom of carbon and three atoms of oxygen.

Since red jellybeans represent sodium atoms, white jellybeans represent hydrogen atoms, black jellybeans represent carbon atoms and blue jellybeans represent oxygen atoms, each of the two students will require the following number of each jellybean for their model of sodium carbonate: One red jellybean, one white jellybean, one black jellybean and three blue jellybeans.

Altogether for both models; two red jellybeans, two white jellybeans, two black jellybeans and six blue jellybeans.

8 0
3 years ago
Calculate the standard entropy change for the following reaction cu(s) + 1/2 O2(g) —&gt; cuo(s)
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3433Explanation:

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8 0
3 years ago
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explain the relationship between the rate of effusion of a gas and its molar mass. methane gas (ch4) effuses 3.4 times faster th
Musya8 [376]

The molar mass of the unknown gas is 184.96 g/mol

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>How to determine the molar mass of the unknown gas </h3>

The following data were obtained from the question:

  • Rate of unknown gas (R₁) = R
  • Rate of CH₄ (R₂) = 3.4R
  • Molar mass of CH₄ (M₂) = 16 g/mol
  • Molar mass of unknown gas (M₁) =?

The molar mass of the unknown gas can be obtained as follow:

R₁/R₂ = √(M₂/M₁)

R / 3.4R = √(16 / M₁)

1 / 3.4 = √(16 / M₁)

Square both side

(1 / 3.4)² = 16 / M₁

Cross multiply

(1 / 3.4)² × M₁ = 16

Divide both side by (1 / 3.4)²

M₁ = 16 / (1 / 3.4)²

M₁ = 184.96 g/mol

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

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3 0
2 years ago
Help me due tomorrow :,)
lesya [120]

#a

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It means

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#c

No reaction because

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#d

Reaction occurs

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Order:-

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3 years ago
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