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3 years ago

7

The image formed by a lens may be real or virtual. The image formed by a lens is always virtual.

2 answers:

amid [387]3 years ago

8 0

There are types of lens; concave and convex lens.
The concave lens is a lens which has an inward curve in the middle, that is, the edges of the curve are thicker than the center of the lens, because of this, any light that enter the lens will spread out [diverge]. An image will look smaller and upright when viewed by a concave lens. Image formed by concave lens are usually VIRTUAL.
A concave lens will produce a real image ONLY if the object is located beyond the focal point of the lens.
A convex lens is a converging lens, this is because, the center of the lens is thicker than its edges. Any ray of light that passes through the lens will converge at the middle of the lens at point called principal focus. A convex lens produce a VIRTUAL image when the object is placed infront of the focal point. The virtual image formed is always magnified and upright.<span />

daser333 [38]3 years ago

5 0

The image formed by a <u>convex</u> lens may be real or virtual. The image formed by a <u>concave</u> lens is always virtual.

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What is polar and unpolar bonds?

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Answer:Nonpolar bonds form between two atoms that share their electrons equally. Polar bonds form when two bonded atoms share electrons unequally.

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What is the atomic mass of an element that has 6 protons, 7 neutrons, and 6 electrons?

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The chemical shift of the CH3 protons in diethylether is δ = 1.16 and that of the CH2 protons is 3.36. What is the difference in

grandymaker [24]

Answer:

Where Blocal = local magnetic field between the two regions of the molecule

Blocal = (1-σ)B0

ΔBlocal = (1-σ1)B0 - (1-σ2)B0 = (σ2 - σ1)B0 = ΔσB0 ≈ ΔδB0 x 10∧-6

= (3.36-1.16) x 10∧-6 x B0 = 2.20 x 10∧-6B0

(a) ΔBlocal = 2.20 x 10∧-6 x 1.9T = 4.2 μT

(b) ΔBlocal = 2.20 x 10∧-6 x 16.5T = 36.3 μT

Explanation:

5 0

3 years ago

The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,

ludmilkaskok [199]

Answer: $4.3\times 10^{-13}s^{-1}$

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $775.0$ = $?$

$Ea$ = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $775.0K$

Now put all the given values in this formula, we get

$\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]$

$\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150$

$(\frac{6.1\times 10^{-8}}{K_2})=141253.8$

Therefore, the value of the rate constant at 775.0 K is $4.3\times 10^{-13}s^{-1}$

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3 years ago