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ivann1987 [24]
4 years ago
9

While developing his periodic table, Dmitri Mendeleev broke his own rule and placed the element tellurium before iodine in the g

roup. He did this even though tellurium's atomic mass was known to be greater than iodine's. In his notes, Mendeleev stated that he expected tellurium to have a lower atomic mass than iodine. However, all measurements continue to show that tellurium has a greater atomic mass than iodine even though tellurium acts like it has a lower atomic mass. What is the most likely reason that Mendeleev placed tellurium before iodine? Mendeleev observed that tellurium has chemical properties like other elements in its group, and he did not know that neutrons cause the greater atomic mass. Iodine isotopes usually have more neutrons than tellurium isotopes, but only the protons affect the properties that Mendeleev could have observed. Mendeleev observed that tellurium and iodine both had very similar chemical properties, but he did not know that iodine atoms have more electrons. Iodine atoms always have more protons than tellurium atoms, but Mendeleev only weighed elements without observing any other properties.
Physics
2 answers:
Evgesh-ka [11]4 years ago
6 0
The answer would be the first option.
Mandeleev observed that tellurium has chemical properties like other elements in its group, and he did not know that neutrons cause the greater atomic mass
svet-max [94.6K]4 years ago
5 0

Answer:

Mendeleev observed that tellurium has chemical properties like other elements in its group, and he did not know that neutrons cause the greater atomic mass.

Explanation:

Neutrons were discovered after Mendeleev's death, so for him order the elements by atomic mass was equivalent to order them by atomic number. It was correct that tellurium goes before iodine, because tellurium has a lower atomic number than iodine, even though it has a greater atomic mass.

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Propane, the gas that is used in barbecue grills, is mad of carbon and hydrogen.Will the atoms that make up propane from covalen
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Now, the atoms of propane are made up of Carbon and hydrogen. These elements are non-metal and thus as described about covalent bonds, they will form covalent bonds because theg are both nonmetals.

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3 years ago
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Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object
vivado [14]

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C. The initial momentum should be equal to the final momentum due to the conservation of momentum.

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Since m/(M+m) < 1, v_1 > v_0.

Explanation:

Wrong -> A. Since the smaller particle still moves after the collision, it has a kinetic energy.

Wrong -> B. The total initial momentum is equal to the momentum of the smaller particle. Therefore, the momentum of the objects that stuck together is equal to that of the smaller object.

Wrong -> D. Since the bigger object is initially at rest and the surface is frictionless, the direction of motion will be the same as the direction of the smaller particle.

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3 years ago
A fighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. On the g
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"Apparent weight during the "plan's turn" is  519.4 N

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The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is \mathrm{a}_{\mathrm{n}}=\frac{v^{2}}{R}

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v = 420 m/s

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a_{n}=\frac{420^{2}}{11000}

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a_{n}=16.03 \mathrm{m} / \mathrm{s}^{2}

It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vectorW_{\mathrm{app}}=m(\text { vector } g \text { -vector a })

In magnitude,

| \text { vector } g-\text { vector } a |=\sqrt{\left(g^{2}+a^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{\left(9.8^{2}+16.03^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{(96.04+256.96)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{353}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=18.78 \mathrm{m} / \mathrm{s}^{2}

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

\mathrm{W}_{\mathrm{app}}=(18.78 \mathrm{m} / \mathrm{s}^ 2)(53 \mathrm{kg})=995.77 \mathrm{N}

Which is quite heavier than his/her true weigh of 519.4 N

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