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3 years ago

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Consider the molecular orbital model of benzene. In the ground state, how many molecular orbital's are filled with electrons? Se

lect one:
A. 1
B. 2
C. 3
D. 4
E. 5
F. 6
G. 7
H. 8

1 answer:

OLga [1]3 years ago

5 0

Answer:

G

Explanation:

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Butanoic acid + 2 propanol

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Answer:

Butanoic acid and 2-propanol reacts to form isopropyl butyrate.

Explanation:

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3 years ago

The orbit closest to the nucleus has ___________ energy.

deff fn [24]

Answer:

inonic bonds with cavalent bonds

Explanation:

ionic bonds

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3 years ago

A 0.4272 g sample of an element contains 2.241 x 10 ^21 atoms . what is the symbol of the element?

grin007 [14]

Answer:

Likely $\rm In$ (indium.)

Explanation:

Number of atoms: $N = 2.241 \times 10^{21}$ .

Dividing, $N$ , the number of atoms by the Avogadro constant, $N_A \approx 6.023 \times 10^{23} \; \rm mol^{-1}$ , would give the number of moles of atoms in this sample:

$\begin{aligned} n &= \frac{N}{N_{A}} \\ &\approx \frac{2.241 \times 10^{21}}{6.023 \times 10^{23}\; \rm mol^{-1}} \approx 3.72 \times 10^{-3}\; \rm mol \end{aligned}$ .

The mass of that many atom is $m = 0.4272\; \rm g$ . Estimate the average mass of one mole of atoms in this sample:

$\begin{aligned}M &= \frac{m}{n} \\ &\approx \frac{0.4272\; \rm g}{3.72 \times 10^{-3}\; \rm mol} \approx 114.82\; \rm g \cdot mol^{-1}\end{aligned}$ .

The average mass of one mole of atoms of an element ( $114.82\; \rm g \cdot mol^{-1}$ in this example) is numerically equal to the average atomic mass of that element. Refer to a modern periodic table and look for the element with average atomic mass $114.82$ . Indium, $\rm In$ , is the closest match.

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3 years ago

Which of the following metals (M) will form an ionic compound with nitrogen with the general formula M3N2?

madreJ [45]

I think it’s a or b i am so not positive

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3 years ago

Read 2 more answers

The solutions that are mixed contain agno3 and nacl. What two products can be formed by exchange of the ions from these two reac

anastassius [24]

Answer:

NaNO₃ and AgCl are the two products that can be formed.

Sodium nitrate, an aqueous solution and a solid silver chloride (precipitate)

Explanation:

We determine the dissociation of both salts

AgNO₃ (aq) → Ag⁺ (aq) + NO₃⁻ (aq)

NaCl (aq) → Na⁺ (aq) + Cl⁻ (aq)

We make the ionic equation:

Ag⁺ (aq) + NO₃⁻ (aq) + Na⁺ (aq) + Cl⁻ (aq) → NaNO₃(aq) + AgCl (s) ↓

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3 years ago