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Rudik [331]
3 years ago
10

The alkali metals all react vigorously with water. Which alkali metal would react to the greatest extent if you have equal amoun

ts of each? A) K B) Li C) Na D) Rb
Physics
1 answer:
Vladimir79 [104]3 years ago
8 0
<h3><u>Answer;</u></h3>

<em>-Rubidium (Rb)</em>

<h3><u>Explanation;</u></h3>
  • <em><u>Alkali metals are group 1 elements, </u></em>they have a valency of 1 and react by loosing electrons to attain a stable configuration.
  • <em><u>The reactivity of alkali metals increases down the group, this is due to the increase in the number of energy levels</u></em> which makes it easier for the valence electrons to be lost.
  • Therefore, in our case,<em><u> rubidium will be the most reactive as it contains more number of energy levels</u></em>, therefore it will loose valence electron more easily compared to the other alkali metals above it in the group.

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What name is given to a process where a substance melts under pressure?​
Digiron [165]

Answer:

Melting

Explanation:

The process of a solid becoming a liquid is called melting. (an older term that you may see sometimes is fusion). The opposite process, a liquid becoming a solid, is called solidification. For any pure substance, the temperature at which melting occurs—known as the melting point—is a characteristic of that substance.

8 0
3 years ago
light of a wavelength 600 nm shines on a soap bubble film. For what soap film thickness will destructive interference occur
VashaNatasha [74]

Answer:

The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

Explanation:

Given;

wavelength of light, λ = 600 nm

The minimum thickness of the soap bubble for destructive interference to occur is given by;

t = \frac{\lambda/n}{2}\\\\t = \frac{\lambda}{2n}

where;

n is refractive index of soap film = 1.33

t = \frac{\lambda}{2n} \\\\t = \frac{600*10^{-9}}{2(1.33)}\\\\t = 2.2556 *10^{-7} \ m\\\\t =  225.56 *10^{-9} \ m\\\\t = 225.56 \ nm

Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

4 0
3 years ago
While escaping from a building in which a fire is burning, 40 people pass through a smoky
kherson [118]

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8 0
3 years ago
Which observation BEST supports evidence that two different species share a common ancestor? A) they both prey on the same anima
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3 0
3 years ago
Read 2 more answers
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

\theta=3.90^{\circ}

Hence, this is the required solution.

8 0
3 years ago
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