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2 years ago

A running track measures 1056 ft per lap. To run 15.0 km, how many laps should you run?

Chemistry

1 answer:

Neporo4naja [7]2 years ago

7 0

Answer:

46.6 laps

Explanation:

Step 1: Given data

Length of the lap (l): 1056 ft

Total distance to be run (d): 15.0 km

Step 2: Convert "l" to km

We will use the conversion factor 1 km = 3280.84 ft.

1056 ft × (1 km / 3280.84 ft) = 0.3219 km

Step 3: Calculate the number of laps that you should run

We will use the following expression.

laps = d/l

laps = 15.0 km/(0.3219 km/lap)

laps = 46.6 laps

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At 10.°C, 20.g of oxygen gas exerts a pressure of 2.1atm in a rigid, 7.0L cylinder. Assuming ideal behavior, if the temperature

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Answer:

Final pressure = 2.3225 atm

Amontons’s law states that

At constant volume and number of molecules, the pressure of a given mass of gas is directly proportional to its temperature

Explanation:

Temperature causes increased excitement of gas molecules increasing the number of collisions with the walls of the container which is sensed as increase in pressure

Amontons’s law: P/T = Constant at constant V and n

That is P1/T1 = P2/T2

Where temperature is given in Kelvin

Hence T1 of 10°C = 273.15 + 10 = 283.15K

Also temperature T2 of 40°C = 313.15 K

Hence

P2 = (P1/T1)×T2 = (2.1/283.15)×313.15 = 2.3225 atm

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2 years ago

12. The most common factors that cause chemical reactions to occur are all the following except

ruslelena [56]

Transfer of electrons

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2 years ago

A compound has the name iodine pentafluoride. What kind of bond is present between the atoms? A. O Metallic B Covalent C. Hydrog

8090 [49]

Answer:

IF(subscript 5)

Explanation:

covalent.

3 0

1 year ago

What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)

Delvig [45]

Explanation:

The given data is as follows.

Energy of radiation absorbed by the electron in hydrogen atom = $1.08 \times 10^{-17} J$

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

E = $h \nu$

$1.08 \times 10^{-17} J$ = $6.626 \times 10^{-34} Js \times \nu$

$\nu$ = $0.163 \times 10^{17} s^{-1}$

or, $\nu$ = $1.63 \times 10^{16} s^{-1}$

It is known that, $\nu = \frac{c}{\lambda}$

$1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}$

$\lambda$ = $1.84 \times 10^{-8} m$

And, according to De-Broglie equation $\lambda = \frac{h}{p}$

as, p = $m \times \nu$

So, $\lambda = \frac{h}{m \times \nu}$

$m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}$

= $3.6 \times 10^{-26} J/m$

Now, on squaring both the sides we get the following.

$(m \times \nu)^{2}$ = $(3.6 \times 10^{-26} J/m)^{2}$

= $12.96 \times 10^{-52}$

$m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

where, m = mass of electron

So, $m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}$

= $\frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}$

= $1.42 \times 10^{-21}$ J

Since, K.E = $\frac{1}{2}m \nu^{2}$

= $\frac{1.42 \times 10^{-21} J}{2}$

= $0.71 \times 10^{-21} J$

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is $7.1 \times 10^{-22} J$ .

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2 years ago

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Answer:

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The particles do not pass through ordinary filter paper.
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The marinade is simply a suspension.
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Also, colloids do not settle on standing.

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2 years ago

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