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Evgesh-ka [11]
2 years ago
13

In the Bohr model of the atom, the electrons move in fixed, ________ paths around a dense, positively-charged nucleus. On the ot

her hand, the quantum mechanical model shows the probability of finding an electron as a ___________ of negative charge.
Question 18 options:

circular; cloud


horizontal; cloud


circular; beam


even, photon
Chemistry
1 answer:
trapecia [35]2 years ago
6 0
Answer:

Answers are in parentheses.

In the Bohr model of the atom, the electrons move in fixed, (circular) paths around a dense positively-charged nucleus. On the other hand, the quantum mechanical model shows the probability of finding an electron as a (cloud) of negative charge.
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2.77mg caffeine / 1oz12oz / 1canLethal dose: 10.0g caffeine = 10,000mg caffeine First, find how much caffeine is in one can of soda, then divide that amount by the lethal dose to find the number of cans. (2.77mg caffeine / 1oz) * (12oz / 1can) = 33.24mg caffeine / 1can. (10,000mg caffeine) * (1can / 33.24mg caffeine) = 300.84 cans. Since we can't buy parts of a can of soda, then we have to round up to 301 cans. Notice how all the values were set up as ratios and how the units cancelled.
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3 years ago
If you wanted to change the polarity of hydrogen bromide (HBr) by substituting
Tcecarenko [31]
Chlorine. Electronegativity generally increases up and across the periodic table
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3 years ago
Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, det
Fed [463]

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² *  3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

2 \times  \frac{4}{3} \pi r^3

(∴ BCC, effective number of atom is 2)

Volume of atoms =

2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3

= 2.55*10⁻²³cm³

\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}

=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24

<h3>therefore, the atomic packing factor of Sn is 0.24</h3>
4 0
3 years ago
What is the greatest source of radiation most humans are exposed to
Scilla [17]
The greatest source of radiation is radon gas
8 0
2 years ago
Write a ground state electron configuration for each neutral atom
Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0
2 years ago
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