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hichkok12 [17]
2 years ago
8

The following molecule can be synthesized from an epoxide and alkyllithium reagent, followed by aqueous workup. However, there a

re two ways in which this molecule can be formed. In one approach, the epoxide has a higher molecular weight than the epoxide used in the other approach. Draw both versions in the boxes below. Draw the carbon-lithium bond as a covalent bond.Part 1:Draw the combination of epoxide and alkyllithium reagent that uses a lower molecular weight epoxide.Part 2: Draw the combination of epoxide and alkyllithium reagent that uses a higher molecular weight epoxide.
Physics
1 answer:
Assoli18 [71]2 years ago
8 0
X. X





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I I. I.

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A 96-mH solenoid inductor is wound on a form 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoi
Sholpan [36]

Answer:

i = 7.83 \mu A

Explanation:

Induced EMF in the coil is given by the equation

EMF = M\frac{di}{dt}

so we have

M = 31 \mu H

also we know that rate of change in current in solenoid is given as

\frac{di}{dt} = 2.5 A/s

so induced EMF of coil is given as

EMF = (31 \times 10^{-6})(2.5)

EMF = 77.5 \times 10^{-6} A/s

now induced current in the coil will be given as

i = \frac{EMF}{R}

i = \frac{77.5 \times 10^{-6}}{9.9}

i = 7.83 \mu A

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
While playing basketball in PE class, Logan lost his balance after making a lay-up and colliding with the padded wall behind the
olga2289 [7]

Answer:

a.) F = 3515 N

b.) F = 140600 N

Explanation: given that the

Mass M = 74kg

Initial velocity U = 7.6 m/s

Time t = 0.16 s

Force F = change in momentum ÷ time

F = (74×7.6)/0.16

F = 3515 N

b.) If Logan had hit the concrete wall moving at the same speed, his momentum would have been reduced to zero in 0.0080 seconds

Change in momentum = 74×7.6 + 74×7.6

Change in momentum = 562.4 + 562.4 = 1124.8 kgm/s

F = 1124.8/0.0080 = 140600 N

6 0
2 years ago
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

            W = ΔK

Where work is defined as the product of force by distance

           W = fr x cos 180

The angle is because the friction force opposes the movement

          Δk =K_{f} –K₀

          ΔK = 0 - ½ m v₀²

We substitute

         - fr x = - ½ m v₀²      

           fr = ½ m v₀²/x

8 0
2 years ago
You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A 2.00-c
AleksAgata [21]

Answer:

Explanation:

Let L be the length of the wire.

velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L  m /s

mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m

m = 7.25 x 10⁻⁷ kg / m

Tension in the wire = Mg  , M is mass hanged from lower end.

= .4 x 9.8

= 3.92 N

expression for velocity of wave in the wire

v = \sqrt{\frac{T}{m} }    , T is tension in the wire , m is mass per unit length of wire .

40.48 L = \sqrt{\frac{3.92}{7.25\times10^{-7}} }

1638.63 L² = 3.92 / (7.25 x 10⁻⁷)

L² = 3.92 x 10⁷ / (7.25 x 1638.63 )

L² = 3299.64

L = 57.44 m /s

5 0
2 years ago
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