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3 years ago

The atom with the same number of neutrons as 12C is 9Be. 11B. 13C. 14N.

Chemistry

1 answer:

julia-pushkina [17]3 years ago

6 0

Answer : The atom with the same number of neutrons as $^{12}C$ is, $^{11}B$

Explanation :

The given atom is, $^{12}C$

Atomic mass number = 12

Atomic number = 6

Atomic number = Number of protons = Number of electrons = 6

Number of neutrons = Atomic mass number - Atomic number = 12 - 6 = 6

Now we have to determine the number of neutrons in the given options.

In atom, $^{9}Be$

Atomic mass number = 9

Atomic number = 4

Atomic number = Number of protons = Number of electrons = 4

Number of neutrons = Atomic mass number - Atomic number = 9 - 4 = 5

In atom, $^{11}B$

Atomic mass number = 11

Atomic number = 5

Atomic number = Number of protons = Number of electrons = 5

Number of neutrons = Atomic mass number - Atomic number = 11 - 5 = 6

In atom, $^{13}C$

Atomic mass number = 13

Atomic number = 6

Atomic number = Number of protons = Number of electrons = 6

Number of neutrons = Atomic mass number - Atomic number = 13 - 6 = 7

In atom, $^{14}N$

Atomic mass number = 14

Atomic number = 7

Atomic number = Number of protons = Number of electrons = 7

Number of neutrons = Atomic mass number - Atomic number = 14 - 7 = 7

Therefore, the atom with the same number of neutrons as $^{12}C$ is, $^{11}B$

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Who is considered the founder of the periodic table of the elements?

nata0808 [166]

Answer:

Dmitry Mendeleev

Explanation:

Around 1869 a Russian scientist, Dmitry Mendeleev formed what is now known as the periodic table or chart. The Mendeleevian periodic table was based on the atomic weights of elements using the periodic law. The periodic law states that "chemical properties of elements are a periodic function of their atomic weights".

The modern periodic table was re-stated by Henry Moseley in the 1900s. He changed the basis of the periodic law to atomic masses.

7 0

3 years ago

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit

natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.

7 0

3 years ago

T/F___ At the eutectic composition, an alloy can solidify at a constant temperature.___ For effective dispersion strengthening,

azamat

Answer:

TRUE
FALSE
TRUE
FALSE
FALSE
TRUE

Explanation:

At the eutectic composition, an alloy can solidify at a constant temperature : TRUE . this is because at eutectic composition the type of reaction that takes place there is invariant reaction in its thermal equilibrium
For effective dispersion strengthening, the dispersed phase should be needle-like, as opposed to round : FALSE. because the rounded shape will not cause a crack.
Intermetallic compounds are usually hard and brittle : TRUE. because Intermetallic compounds prevents dislocation movements and this makes them brittle and hard
For the effective dispersion and strengthening, the dispersed phase should be continuous : FALSE. this is because the dispersed precipitate must be small and not continuous
Stoichiometric intermetallic compounds exist over a range of compositions : FALSE
Faster solidification results in smaller interlamellar spacing : TRUE