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3 years ago

At STP, what is the volume of 5.13 mol of nitrogen gas? Answer in units of L.

Chemistry

1 answer:

Gre4nikov [31]3 years ago

7 0

Answer: $114.91L$

Explanation:

You need to know the conversion factor first in order to solve this. Any gas occupies 22.4L per mol.

$5.13mol(\frac{22.4L}{1mol})= 114.91L$ of nitrogen gas.

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Exactly 10.7 mL of water at 28.0 °C is added to a hot iron skillet. All of the water is converted into steam at 100.0°C. The mas

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3 years ago

in an experiment 3.425g of lead oxide was reduced to form 3.105g of lead the empirical formula of the lead oxide is?

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Answer:

Pb3O4

Explanation:

According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,

Mass of lead oxide = 3.425g

Mass of lead = 3.105g

Mass of oxygen = (3.425g - 3.105g) = 0.320g

Next, we convert each mass value to mole by dividing by respective molar mass

Pb = 3.105g ÷ 207.2 = 0.0149mol

O = 0.320g ÷ 16 = 0.02mol

Next, we divide each mole value by the smallest (0.0149)

Pb = 0.0149mol ÷ 0.0149mol = 1

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Multiply each ratio value by 3 to get:

Pb = 1 × 3 = 3

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The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.

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3 years ago

An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average

Deffense [45]

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = $\frac{Displacement }{Change in time}$ = Δx / Δt = $\frac{x2 - x1}{t2 - t1}$

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = $\frac{Distance traveled}{Time taken}$

Time = $\frac{Distance traveled}{Velocity}$

Therefore, Time = $\frac{2100km }{1000km/h}$

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = $\frac{Distance traveled}{Time taken}$

Time = $\frac{Distance traveled}{Velocity}$

Therefore, Time = $\frac{1300km }{800km/h}$

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = $\frac{x2 - x1}{t2 - t1}$

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = $\frac{1300 - 2100}{1.625 - 2.1}$

Average velocity = 1684.21 km/h

7 0

3 years ago