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Zarrin [17]
4 years ago
5

Select the appropriate statements and arrange them in a logical order to explain the bonding in BeF2. i. Each of the singly occu

pied hybrid orbitals on Be overlaps with a singly occupied 2p orbital on an F atom. ii. The unhybridized 2p orbitals on Be interact with the 2p orbitals on the F atoms to form pi bonds. iii. The 2s and one of the 2p orbitals on Be combine to form two sp hybrid orbitals. iv. A 2s electron is promoted to a 2p orbital on Be. v. The 2s electrons in Be are promoted to 2p orbitals. vi. Each of the singly occupied 2p orbitals on Be overlaps with a singly occupied 2s orbital on a F atom.a.v, vi, ii b.v,vi c.iii, i, ii d.iv, iii, i e.iv, iii, i, ii
Chemistry
1 answer:
galina1969 [7]4 years ago
4 0

Answer:

e.iv, iii, i, ii

Explanation:

Be has electron configuration of 1s^2 2s^2. During bond formation, one 2s electron is promoted to a 2p orbital. Hence two hybrid sp beryllium orbitals are formed. These two hybrid orbitals are used in sigma bond formation to fluorine atoms.

Recall that fluorine has filled 2p orbitals containing lone pairs of electrons. These filled orbitals can interact with the unhybridized 2p-orbitals on beryllium to form pi bonds. This is the reason for the sequence of events chosen in the answer.

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6 0
3 years ago
Intact cells of two unknown cell types were placed into solutions with different concentrations of NaCl. Type I cells swelled an
Bess [88]

Answer:

Cell Type I: Animal cell surrounded by a plasma membrane only.

Cell Type II: Plant cell surrounded by a plasma membrane and a cell wall.

Explanation:

Hello,

In this case, we must remember that animal cells are covered only by a plasma membrane, which is not enough to keep the cell from bursting as no limit for the inlet salt is established, causing it both to swell and burst. On the other hand, plant cells are covered by both a plasma membrane and a cell wall which contributive effect allow the inlet of the salt but prevent the cell to burst as the cell wall is rigid. In such a way, based on the described situation, one infers that the cell type I is an animal cell surrounded by a plasma membrane only the cell type II is a plant cell surrounded by a plasma membrane and a cell wall.

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3 years ago
I break and shatter easily and I am not shiny. What am I?
Troyanec [42]

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7 0
2 years ago
During a period of discharge of a lead-acid battery, 405 g of Pb from the anode is converted into PbSO4 (s).
Alexus [3.1K]

Answer:

The answers to the question are as follows

First part

The mass of PbO2 (s) reduced at the cathode during the period is = 467.55_g

Second part

The electrical charge are transferred from Pb to PbO2 is 377186.86_C or 3.909 F  

Explanation:

To solve this, we write the equation for the discharge of the lead acid battery as

H₂SO₄ → H⁺ + HSO₄⁻

Pb (s) + HSO⁻₄ → PbSO₄ + H⁺ + 2e⁻

at the cathode we have

PbO₂ + 3H⁺ + HSO⁻₄ + 2e⁻ → PbSO₄ + 2H₂O

Summing the two equation or the total equation for discharge is

Pb (s) + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

From the above one mole of lead and one mole of PbO₂  are consumed simultaneously hence

Number of moles of lead contained in 405 g of Pb with molar mass  = 207.2 g/mole = (405 g)/ (207.2 g/mole) = 1.95 mole of Pb

Hence number of moles of  PbO₂ reduced at the cathode = 1.95 mole

mass of  PbO₂ reduced at the cathode = (number of moles)×(molar mass)

= 1.95 mole × 239.2 g/mol = 467.55 g of Lead (IV) Oxide is reduced at the cathode

Part B

Each mole of Pb transfers 2e⁻ or 2 electrons, therefore 1.95 moles of Pb will transfer 2 × 1.95 = 3.909 moles of electrons transferred

Each electron carries a charge equal to -1.602 × 10⁻¹⁹ C or one mole of electrons carry a charge equal to 96,485.33 coulombs

hence 3.909 moles carries a charge = 3.909 × 96,485.33 coulombs =377186.86 Coulombs of electrical charge

or transferred electrical charge = 377186.86 C or 3.909 Faraday

6 0
3 years ago
Definition of phosphoric acid
miss Akunina [59]
<em>Phosphoric acid is an acid used in fertilizers and soaps.</em>

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8 0
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