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Levart [38]
2 years ago
12

Classify pH values of 9.1, 1.2, and 5.7 as basic, acidic, or very acidic.

Physics
1 answer:
kramer2 years ago
4 0
9.1 basic since > 7
1.2 VERY acidic << 7
<span>5.7 acidic < 7 </span>
You might be interested in
The wavelength of green light is 550 nm.
SIZIF [17.4K]

Answer:

(a) momentum of photon is 1.205 x 10⁻²⁷ kgm/s

    velocity of electron is 1323.88 m/s

   momentum of the electron is 1.205 x 10⁻²⁷ kgm/s

(b) momentum of photon is 1.506 x 10⁻²⁷ kgm/s

  velocity of electron is 1654.85 m/s

  momentum of the electron is 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron

Explanation:

(a)

wavelength of green light, λ = 550 nm

momentum of photon is given by;

p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{550*10^{-9}}\\\\p = 1.205 *10^{-27} \ kg.m/s

velocity of electron is given by;

P = \frac{h}{\lambda} \\\\mv = \frac{h}{\lambda}\\\\v = \frac{h}{m \lambda}\\\\v =   \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(550*10^{-9})}\\\\v = 1323.88 \ m/s

momentum of the electron is given by;

p = mv

p = (9.1 x 10⁻³¹) (1323.88)

p = 1.205 x 10⁻²⁷ kgm/s

(b)

wavelength of red light, λ = 440 nm

momentum of photon is given by;

p = \frac{h}{\lambda}\\\\ p = \frac{6.626 *10^{-34}}{440*10^{-9}}\\\\p = 1.506 *10^{-27} \ kg.m/s

velocity of electron is given by;

v =   \frac{6.626 *10^{-34}}{(9.1*10^{-31} )(440*10^{-9})}\\\\v = 1654.85 \ m/s

momentum of the electron is given by;

p = mv

p =  (9.1 x 10⁻³¹) (1654.85)

p = 1.506 x 10⁻²⁷ kgm/s

(c) The momentum of the photon is equal to the momentum of the electron.

7 0
2 years ago
A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 t. what is the maximum hall voltage
Alex787 [66]
The equation for Hall voltage Vh is:

Vh=v*B*w, where v is the velocity of the strip, B is the magnitude of the magnetic field, and w is the width of the strip. 

v=25 cm/s = 0.25 m/s
B=5.6 T
w= 1.2 mm = 0.0012 m

We input the numbers into the equation and get:

Vh= 0.25*5.6*0.0012 = 0.00168 V

The maximum Hall voltage is Vh= 0.00168 V.
4 0
2 years ago
Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,
Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(c) 4, 3, 0, 0

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number                                   Symbol                Possible values

Principal Quantum Number  .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l  

Spin Quantum Number.................m_s.....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

(c) Here we have

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l)  → acceptable

The spin quantum number m_s = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

3 0
2 years ago
Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights
Arlecino [84]

Answer:

               275 kPa

Explanation:

             mass of the gas=m=1.5 kg

             initial volume if the gas=V₁=0.04 m³

             initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

             V₂=final volume

             V₂=2 V₁

As the temperature is constant

             T₁=T₂=T

\frac{P1V1}{T1}=\frac{P2 V2}{T2}

putting the values in the equation.

\frac{P1V1}{T1}=\frac{P2 *2V1}{T2}

P₂=\frac{P1}{2}

P₂=\frac{550}{2}

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

           

3 0
2 years ago
Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr
Aloiza [94]

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

F=ma

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0
2 years ago
Read 2 more answers
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