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labwork [276]
4 years ago
5

A mountaineer shouts for help Half a second later (0.5 she hears the echo How far away is the rock face which is reflecting her

voice (take the speed of sound to be 330 m\s) help please I will mark brain liest
Chemistry
1 answer:
irinina [24]4 years ago
7 0

Answer:

THE DISTANCE OF THE ROCK REFLECTING HER VOICE IS 82.5 METERS.

Explanation:

In sound waves, the speed of a sound is equal to twice the distance divided by time.

Mathematically,

Speed = 2 * Distance/ Time

Speed = 33- m/s

Time = 0.5 second

Distance = unknown

So re-arranging and introducing the variables into the formula, we have;

Distance = speed * time / 2

Distance = 330 * 0.5 / 2

Distance = 82.5 meters.

The distance away from the rock face which is reflecting her voice is 82.5 meters.

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Please Answer Correctly
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Explanation: The radius of an object is found from the center of the object to the perimeter. Radius can be any number, but it is the measurement from the center to the perimeter.

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Given that it requires 27.9 ml of 0.270 m na2s2o3(aq) to titrate a 15.0-ml sample of i3–(aq), calculate the molarity of i3–(aq)
Digiron [165]

The molar concentration of the KI_3 solution is 0.251 mol/L.

<em>Step 1</em>. Write the <em>balanced chemical equation</em>

I_3^(-) + 2S_2O_3^(2-) → 3I^(-) + S_4O_6^(2-)

<em>Step 2</em>. Calculate the <em>moles of S_2O_3^(2-)</em>

Moles of S_2O_3^(2-)

= 27.9 mL S_2O_3^(2-) ×[0.270 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)]

= 7.533 mmol S_2O_3^(2-)

<em>Step 3</em>. Calculate the <em>moles of I_3^(-) </em>

Moles of I_3^(-) = 7.533 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 3.766 mmol I_3^(-)

<em>Step 4</em>. Calculate the <em>molar concentration of the I_3^(-) </em>

<em>c</em> = "moles"/"litres" = 3.766 mmol/15.0 mL = 0.251 mol/L

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4 years ago
Which acid(s) is weaker than benzoic acid?
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Explanation : just is
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4 years ago
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Calculate the average molecular weight of air (1) from its approximate molar composition of 79% N2, 21% 02x
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Answer:

Average molecular weight of air is 28.84 g/mol.

Explanation:

The average molecular weight of a mixture is determined from their molar composition and molecular weight.

Average molecular weight :\sum (\chi_i\times m_i)

\chi_1 : mole fraction of the 'i' component.

m_i = Molecular weight of i component

Average molecular weight of air with approximate molar composition of 79% nitrogen gas and 21% of oxygen gas can be calculated as:

Average molecular weight of air:

79\%\times 28 g/mol+21\%\times 32 g/mol

=0.79\times 28 g/mol+0.21\times 32 g/mol=28.84 g/mol

8 0
3 years ago
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