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marta [7]
3 years ago
12

A spring running horizontally has areas with the coils closer together and areas with them farther apart. A label A points to ar

eas where they are closer together. A label B points to areas where they are farther apart. A label C sits near a bracket connecting the center of two successive areas with the coils farther apart, which are separated by an area of closer coils.
Label the parts of the longitudinal wave.

Compressions:

Rarefactions:

Wavelength:
Chemistry
1 answer:
Margaret [11]3 years ago
5 0

Answer:

compressions: a

rarefactions: b

wavelength: c

Explanation:

i got it right

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A sample of methane collected when the temp was 30 C mmHg measures 398 mL. What would be the volume of the sample at -5 C and 61
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<h2>Question </h2>

A sample of methane collected when the temp was 30 C and 760mmHg measures 398 mL. What would be the volume of the sample at -5 C and 616 mmHg pressure

<h2>Answer:</h2>

434.32mL

<h2>Explanation:</h2>

Using the combined gas law:

\frac{PV}{T} = k

Where;

P = Pressure

V = Volume

T = Temperature

k = constant.

It can be deduced that:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = k           ---------------------(i)

Where:

P₁ and P₂ are the initial and final pressures of the given gas

V₁ and V₂ are the initial and final volumes of the given gas

T₁ and T₂ are the initial and final temperatures of the gas.

<em>From the question:</em>

the gas is methane

P₁  = 760mmHg

P₂ = 616mmHg

V₁ = 398mL

V₂ = ?

T₁ = 30°C = (30 +273)K = 303K

T₂ = -5°C = (-5 +273)K = 268K

Substitute these values into equation (i) as follows;

\frac{760*398}{303} = \frac{616*V_2}{268}

Solve for V₂

V₂ = \frac{760*398*268}{616*303}

V₂ = 434.32mL

Therefore, the volume of the sample at -5C and 616mmHg pressure is 434.32mL

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