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3 years ago

What is oxygen an example of? a mixture, an element, a compound, or a solution

Physics

1 answer:

soldi70 [24.7K]3 years ago

7 0

Answer: element

Explanation:

Couldn't be a compound because contains same type of atoms. Can't be a mixture either because not physically mixed or anything. Isn't a solution ofc because oxygen is a gas.

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A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle

alexira [117]

<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>

According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.

Therefore; Sin i/Sin r = η

In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

= 0.3759

r = Sin^-1 0.3759

= 22.08

<u>≈ 22°</u>

3 0

3 years ago

Which branches of natural science include the study of an organism that lived 10 million years ago?

Verdich [7]

Analyzing the components and decay products of school lunches.

7 0

3 years ago

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A gas takes up a volume of 17.0 L, has a pressure of 2.30 atm, and a temperature of 299 K. If the temperature is raised to 350.0

Maurinko [17]

Answer:

new volume = 30.513L

Explanation:

V1 =17L , P1= 2.3atm , T1=299K , T2=350K , P2 =1.5atm, V2=?

To find the new volume of the gas we will use the combined gas law.

The combined gas law combines Boyle's law, Charles law and Gay-Lussac's law. Its states that the ratio of the product of pressure and volume to temperature is equal to a constant. its expressed mathematically below.

$\frac{V1P1}{T1} = \frac{V2P2}{T2}$

$V2 =\frac{V1P1T2}{P2T1}$

$V2 =\frac{17*2.3*350}{1.5*299}$

$v2 = \frac{13685}{448.5}$

V2 =30.513L

4 0

3 years ago

Read 2 more answers

The electric field intensity between two large,

Ivahew [28]

Answer:

300 is the answer

Explanation:

Hope that this answer will help you

8 0

3 years ago

An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V batte

Lelechka [254]

Answer:

$2.25\mu C$

Explanation:

At the beginning, we have:

V = 4.0 V potential difference across the capacitor

$Q=9.0 \mu C=9.0\cdot 10^{-6}C$ charge stored on the capacitor

Therefore, we can calculate the capacitance of the capacitor:

$C=\frac{Q}{V}=\frac{9.0 \cdot 10^{-6} C}{4.0 V}=2.25\cdot 10^{-6} F$

Later, the battery is replaced with another battery whose voltage is

V = 5.0 V

Since the capacitance of the capacitor does not change, we can calculate the new charge stored:

$Q=CV=(2.25\cdot 10^{-6} F)(5.0 V)=11.25 \cdot 10^{-6} C=11.25 \mu C$

Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from $9.0 \mu C$ to $11.25 \mu C$ . Therefore, the additional charge that moved to the positive plate is

$\Delta Q = 11.25 \mu C-9.0 \mu C=2.25 \mu C$

5 0

3 years ago