Mendeleev gave the name eka-aluminum to an A. compound containing aluminum. B. mixture of aluminum and an unknown element. C. un
known element he predicted would have properties similar to those of aluminum. D. rare isotope of aluminum.
2 answers:
Mendeleev gave the name eka-aluminum to an <u><em>C. unknown element he predicted would have properties similar to those of aluminum </em></u>
<h3>Further explanation </h3>The periodic system of elements is an arrangement of elements classified based on similar properties and their atomic number order.
The periodic system is divided into groups (vertical order) and periods (horizontal order)
There are some scientists who do the grouping of elements, mostly based on the relative atomic mass
- 1. Dobereiner
These scientists grouped the elements into groups of three elements called the Triade
- 2. Newlands
These scientists arrange elements based on their relative atomic mass increases called the octave law. Similarities between the elements occur in elements that have a difference of 1 octave (elements 1 and 8, elements 2 and 9, elements 3 and 10 etc.)
- 3. Mendeleev
These scientists group elements based on their relative atomic mass increases.
Mendeleev placed elements that have similar properties in one group (vertical direction) and based on the increase in relative atomic mass in one period
In the periodic system made by Mendeleev, there are some parts in the group that are left blank because the element has not been found but is predicted to have the same properties as the elements in that group.
The Mendeleev periodic system is useful after the discovery of new elements which had similar properties in accordance with Mendeleev's predictions.
In group III, under the Al element, Mendeleev empties the unknown element, but it is predicted to have similar properties with aluminum and is given the name eka-aluminum. This element was later discovered and named gallium (Ga) with atomic number 31 and relative atomic mass of 69.72
<h3>Learn more </h3>
electron affinity
Identify the group number in the periodic table
the charge on each ion in the compounds
Keywords: element, group, period, eka-aluminum, gallium, periodic system
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This question is incomplete, the complete question;
you make an interferometer using 50-50 beam splitter and two mirrors, one being a perfect mirror and one which does not reflect all light. The wavelength of the 9 mW incident laser is 400 nm.
Because the top mirror is not perfectly reflective (it reflects 90% of the photons, allowing 10% of them to go through), the power measured at the detector when only the vertical arm is blocked is 2.25 mW, while the power measured at the detector when only the horizontal arm is blocked is only 2.025 mW. Assume initially the intensity is at its maximum. How much would we need to translate the perfect mirror to the right to get a minimum intensity at detector, and what is that minimum intensity
Options;
a) 200 nm; 0.9 mW
b) 100 nm, 0.0059 mW
c) 200 nm; 0 mW
d) 100 nm; 0.9 mW
e) 200 nm; 0.0059 mW
Answer:
the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer
Explanation:
Given that the instrument here is an interferometer.
Maximum intensity is obtained when the two waves are exactly in phase.
that is the peaks (crusts and troughs) and nodes (zero value points) of the two waves will be at the exact same point when the wave falls on the detector.
The phase factor of this point is taken as ∅ = 0
Now, to get a minimum point, the phase difference between the two waves should be should be ∅ = π
This corresponds to a path difference between the two waves as half of the wavelength. λ/2
The light gets reflected from the mirror.
Hence, when we move the mirror by a length l, the extra/less path the light has to travel is 2l (light is going and coming back)
hence, to get a path difference of λ/2 the mirror should move half of this distance only
so, the mirror should move;
= λ/4
here, wavelength is 400nm
the length moved by the mirror = 400/4 = 100 nm
The intensity is given by the equation;
=
1 +
2 + 2√
1
2cos(∅)
where
1 = 2.25 mW
2 = 2.025 mW
∅ = π
so we substitute
= 2.25 + 2.025 - 2√(2.25 × 2.025)
= 4.275 - 4.26907
= 0.0059
Therefore; the amount we need to translate the perfect mirror to the right to get a minimum intensity at detector and the minimum intensity are;
100 nm; 0.0059 mW
Option b) 100 nm, 0.0059 mW is the correct answer
Answer:
The white car will cover the most distance every second.
Explanation:
The formula for the uniform speed of an object is given as follows:
where,
s = distance covered by the object
v = speed of the object
t = time required
Now, if we assume time to be constant at 1 s. Then the distance covered by each car will be directly proportional to the speed of the car. Hence, the car with the greatest speed will travel the greatest distance in 1 second.
We, have:
Speed of white car > Speed of red car > Speed of green car
<u>Therefore, the white car will cover the most distance every second.</u>