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coldgirl [10]
3 years ago
6

(WORTH 40 PTS.) Are the compounds in Figure 21-3 substituted hydrocarbons? How do you know?

Chemistry
1 answer:
maks197457 [2]3 years ago
6 0

Answer:

Yes

Explanation:

The compounds have both hydrogen and carbon only which are atoms that make up the hydrocarbons and they are also gases at room temperature

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Which of the following bonds will have the smallest difference in electronegativity?
Vitek1552 [10]

Answer:

The answer to your question is the first choice (H₂)

Explanation:

Process

Look for the electronegativity of the elements of this exercise

a) H₂ = 2.2 - 2.2 = 0

b) NaF = 3.98 - 0.93 = 3.05

c) HBr = 2.96 - 2.2 = 0.76

d) HS⁻ = 2.58 - 2.2 = 0.38

The molecule that has the smallest difference in electronegativity is H₂

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Answer:

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Explanation:

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3 years ago
The force known as the collisions of gas particles of matter is known as __________.
bazaltina [42]

Answer: the answer is combustion

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4 years ago
The temperature of the shadows of your increases with altitude because of the presence of?
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is the answer i believe

Explanation:

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3 years ago
Part B Identify the sets of quantum numbers that describe all the electrons in the ground state of a neutral beryllium atom, Be.
dmitriy555 [2]

<u>Answer:</u> The set of quantum numbers for the electrons in Be atom are (2, 1, -1, 1/2), (1, 0, 0, 1/2), (1, 0, 0, -1/2) and (2, 0, 0, -1/2)

<u>Explanation:</u>

There are 4 quantum numbers:

  • Principal Quantum number (n) specifies the energy of the electron in a shell.
  • Azimuthal Quantum number (l) specifies the shape of an orbital. The value of it lies in the range of 0 to (n-1)
  • Magnetic Quantum number (m) specifies the orientation of the orbital in space. The value of it lies in the range of -l to +l
  • Spin Quantum number (s) specifies the spin of an electron in an orbital. It can either have a value of +\frac{1}{2} or -\frac{1}{2}

Berylium (Be) is the 4th element of periodic table having electronic configuration of 1s^22s^2

  • <u>For electrons in 1s-orbital, the quantum numbers can be:</u>

For first electron:

n=1\\l=0\text{ (for s-subshell)}\\m=0\\s=+\frac{1}{2}

For second electron:

n=1\\l=0\\m=0\\s=-\frac{1}{2}

  • <u>For electrons in 2s-orbital, the quantum numbers can be:</u>

For first electron:

n=2\\l=0\text{ (for s-subshell)}\\m=0\\s=+\frac{1}{2}

For second electron:

n=2\\l=0\text{ (for s-subshell)}\\m=0\\s=-\frac{1}{2}

Hence, the set of quantum numbers for the electrons in Be atom are (2, 1, -1, 1/2), (1, 0, 0, 1/2), (1, 0, 0, -1/2) and (2, 0, 0, -1/2)

4 0
3 years ago
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