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2 years ago

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13. Which compound is most reactive? ethene ethyne ethane

2 answers:

Deffense [45]2 years ago

6 0

Answer: Ethyne

Explanation:

katen-ka-za [31]2 years ago

4 0

Answer:

Ethyne consists of a triple bond that is most easier to be broken as compared to that of ethyne with a double bond and benzene single bonds. Therefore ethyne is most reactive among the given options.

Explanation:

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Answer:

C

Explanation:

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3 years ago

Which weather event is capable of destroying homes and uprooting trees due to a low pressure area at its center

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3 years ago

Read 2 more answers

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

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3 years ago

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andre [41]

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2 years ago

Assuming that the limits of the visible spectrum are approximately 380 and 700 nm, find the angular range of the first-order vis

ch4aika [34]

Answer:

angular range is ( 0.681 rad , 0.35 rad )

Explanation:

given data

wavelength λ = 380 nm = 380 × $10^{-9}$ m

wavelength λ = 700 nm = 700 × $10^{-9}$ m

to find out

angular range of the first-order

solution

we will apply here slit experiment equation that is

d sinθ = m λ ...........1

here m is 1 for single slit and d is = $\frac{1}{900*10^3 m}$

so put here value in equation 1 for 380 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 380 × $10^{-9}$

θ = 0.35 rad

and for 700 nm

we get

d sinθ = m λ

$\frac{1}{900*10^3}$ sinθ = 1 × 700 × $10^{-9}$

θ = 0.681 rad

so angular range is ( 0.681 rad , 0.35 rad )

3 0

3 years ago