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3 years ago

Could a mixture be made up of only elements ?

1 answer:

3 0

Absolutely ! You're breathing a mixture like that right now.

Earth's atmosphere ... except for the small amounts of carbon dioxide
and water vapor ... is a mixture of gases in their elemental form. A few
of them include nitrogen, oxygen, neon, argon, krypton. All elements.

You might be interested in

Can u guys check my 3 anwsers? *^* And tell me if i got them right?

viva [34]

My answers would be as follows:

<span>You step on the scale and notice that you have lost five pounds. Which of the following has changed?

B) Only your weight changed since mass is conserved. It cannot be destroyed or created.

How could you increase the mass of a wooden block?

D) None of the above. Mass cannot be created so the mass will stay the same it will only be the weight you can increase depending on the acceleration.

If you went to the moon, how would your mass change relative to your mass on Earth?

A) It would be the same on the moon as on Earth. Mass will be the same. It is weight that will change since the gravitational acceleration in the moon is different as that to the earth.
</span>

8 0

3 years ago

What is the frequency and wavelength, in nanometers, of photons capable of just ionizing nitrogen atoms?

nika2105 [10]

Answer:

The frecuency and wavelength of a photon capable to ionize the nitrogen atom are ν = 3.394×10¹⁵ s⁻¹ and λ = 88.31 nm.

Explanation:It is possible to know what are the frequency and wavelength of a photon capable to ionize the nitrogen atom using the equation of the energy of a photon described below.

E = hc/λ (1)

Where h is the Planck constant, c is the speed of light and λ is the wavelength of the photon.

But first, it is neccesary to know the ionization energy of the nitrogen atom. The ionization energy is the energy needed to remove an electron from an atom, for the Nitrogen atom it will lose an electron of its outer orbit from the nucleus, farther snuff, so the electric force is weaker. Experimentally, it is known that it has a value of 14.04 eV. This value is easy to found in a periodic table.

So the nitrogen atom will need a photon with the energy of 14.04 eV to remove the electron from its outer orbit.

Replacing the Planck constant, the speed of light and the energy of the photon in the equation 1, the wavelength can be calculated:

λ = hc/E (2)

Where h = 6.626×10⁻³⁴ J.s and c = 3.00×10⁸ m/s

But the Planck constant can be expressed in electron volts:

1 eV = 1.602 x 10⁻¹⁹ J

h = 6.626x10⁻³⁴ J/1.602x10⁻¹⁹ J . eV .s

h= 4.136x10⁻¹⁵ eV.s

Now, it is convenient to express the speed of light in nanometers:

1nm = 1x10⁻⁹ m

c = 3.00x10⁸ m/ 1x10⁻⁹ m

c = 3x10¹⁷ nm/s

Substituting in equation 2:

λ = (4.136x10⁻¹⁵ eV.s)(3x10¹⁷ nm/s)/14.04 eV

λ = 1240 eV. nm/ 14.04 eV

λ = 88.31 nm

The frenquency is calculated using the equation 2 in the following way:

E = hν (3)

Where ν is the frecuency

ν = E/h

ν = 14.04 eV/4.136×10⁻¹⁵ eV.s

ν = 3.394×10¹⁵ s-1

So the frecuency of a photon, capable to ionize the nitrogen atom, will be 3.394×10¹⁵ s⁻¹ and its wavelength 88.31 nm.

4 0

4 years ago

when the elevator is moving upwards but its slowing to a stop at the top floor. what is the direction of the net force on the ri

Zielflug [23.3K]

Answer:Downward

Explanation:

As elevator is moving upward but is decelerating to stop at top

Suppose mg is the weight of man

N is the normal reaction

a is the acceleration

$F_{net}=N-mg=m(-a)$

here upward direction is positive while downward direction is negative

Therefore net force is in downward direction

8 0

4 years ago

A train leaves the station heading west on the tracks. It takes the train 8 seconds to reach 80 miles per hour. It completes the

Darina [25.2K]

im not sure about most of this but ill try to help

so there are three formulas you need to know first: the formula for rate distance and time.

to find distance:
distance = rate * time

to find time:
time = distance / rate

and to find rate (which is what we need)
rate = distance / time

now we just use our formula and plug in the numbers

rate = distance * time

rate = 160 / 2 = 80

so our rate is 80. thats all i know i have no idea about the rest but i always use these messurments. im sure you can google most of it to get other formulas and calculators though. sorry i couldnt help much! good luck :D

8 0

4 years ago

Read 2 more answers

I’M DESPERATE PLZ HELP!! 40 POINTS!!

Ierofanga [76]

Answer:

a. The acceleration of the hockey puck is -0.125 m/s².

b. The kinetic frictional force needed is 0.0625 N

c. The coefficient of friction between the ice and puck, is approximately 0.012755

d. The acceleration is -0.125 m/s²

The frictional force is 0.125 N

The coefficient of friction is approximately 0.012755

Explanation:

a. The given parameters are;

The mass of the hockey puck, m = 0.5 kg

The starting velocity of the hockey puck, u = 5 m/s

The distance the puck slides and slows for, s = 100 meters

The acceleration of the hockey as it slides and slows and stops, a = Constant acceleration

The velocity of the hockey puck after motion, v = 0 m/s

The acceleration of the hockey puck is obtained from the kinematic equation of motion as follows;

v² = u² + 2·a·s

Therefore, by substituting the known values, we have;

0² = 5² + 2 × a × 100

-(5²) = 2 × a × 100

-25 = 200·a

a = -25/200 = -0.125

The constant acceleration of the hockey puck, a = -0.125 m/s².

b. The kinetic frictional force, $F_k$ , required is given by the formula, F = m × a,

From which we have;

$F_k$ = 0.5 × 0.125 = 0.0625 N

The kinetic frictional force required, $F_k$ = 0.0625 N

c. The coefficient of friction between the ice and puck, $\mu_k$ , is given from the equation for the kinetic friction force as follows;

$F_k = \mu_k \times Normal \ force \ of \ hockey \ puck = \mu_k \times 0.5 \times 9.8$

$\mu_k = \dfrac{0.0625}{0.5 \times 9.8} \approx 0.012755$

The coefficient of friction between the ice and puck, $\mu_k$ ≈ 0.012755

d. When the mass of the hockey puck is 1 kg, we have;

Given that the coefficient of friction is constant, we have;

The frictional force $F_k = 0.012755 \times 1 \times 9.8 = 0.125 \ N$

The acceleration, a = $F_k$ /m = 0.125/1 = 0.125 m/s², therefore, the magnitude of the acceleration remains the same and given that the hockey puck slows, the acceleration is -0.125 m/s² as in part a

The frictional force as calculated here, $F_k = 0.125 \ N$

The coefficient of friction $\mu_k$ ≈ 0.012755 is constant

5 0

3 years ago