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Komok [63]
3 years ago
10

What is the hybridization of the central iodine atom in i3−?

Chemistry
2 answers:
Ainat [17]3 years ago
8 0
The central iodine atom in triiodide has sp3d hybridization.In triiodide anion, the central iodine atom has three equatorial lone pairs of electrons and the terminal iodines are bonded axially in a linear shape. Electrons in sp3d hybridization are arranged in trigonal bipyramidal symmetry.
Explanation:

When you draw the Lewis structure of this particle, you'll realize that the central I atom has a pair of bonds and three individual pairs of electrons. as a result of there are five things around that central I atom, it's<span> sp3d hybridized.
</span>
The bonds during a gas<span> (CH4) molecule </span>are fashioned<span> by four separate </span>however<span> equivalent orbitals; </span>one<span> 2s and </span>3<span> 2p orbitals of the carbon </span>interbreed<span> into four sp3 orbitals. </span>within the<span> ammonia molecule (NH3), 2s and 2p orbitals </span>produce<span> four sp3hybrid orbitals, </span>one among that<span> is occupied by a lone </span>try<span> of electrons.</span><span>

</span>
Ilya [14]3 years ago
8 0
Answer is: the central iodine atom in triiodide has sp3d hybridization.
In triiodide anion the central iodine atom has three equatorial lone pairs of electrons and the terminal iodines are bonded axially in a linear shape. Electrons in sp3d hybridization <span>are arranged in trigonal bipyramidal symmetry.</span>
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What is the total mass in grams of 0. 75 mole of so2.
Bas_tet [7]

Answer:

48.075g(or 48g in correct sig figs)***

Explanation:

\frac{0.75mol}{1} *\frac{64.1g}{1 mol}=48.075g

*64.1g is the mass of SO2 which is calculated by simply taking the mass of sulfur and oxygen(but doubling it since there are two oxygens) and adding them together(32.1+2x16.0)

**btw the mol units cancel because of dimensional analysis in case anyone was wondering why

***if your teacher is like mine and specifically wants your answer in correct sig figs, use the answer in parentheses as the original problem only has 2 sig figs

7 0
2 years ago
A rigid cylinder with a movable piston contains a sample of hydrogen gas. At 330. K, this sample has a pressure of 150. kPa and
Marta_Voda [28]

Answer:

V₂ = 4.34 L

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 3.50 L

Initial pressure = 150 Kpa (150/101.325 = 1.5 atm)

Initial temperature = 330 K

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

V₂ = P₁V₁ T₂/ T₁ P₂

V₂ =  1.5 atm ×  3.50 L × 273 K / 330 K × 1 atm

V₂ = 1433.3 atm .L. K / 330 k.atm

V₂ = 4.34 L

4 0
3 years ago
How do you write 400000 in scientific notation?
Andrej [43]

Answer:

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Explanation:

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7 0
2 years ago
A solution is 40.00% by volume benzene (C6H6) in carbon tetrachloride at 20°C. The vapor pressure of pure benzene at this temper
finlep [7]

Answer:

The total vapor pressure is 84.29 mmHg

Explanation:

Step 1:  Data given

Solution = 40.00 (v/v) % benzene in CCl4

Temperature = 20.00 °C

The vapor pressure of pure benzene at 20.00 °C = 74.61 mmHg

Density of benzene is 0.87865 g/cm3

The vapor pressure of pure carbon tetrachloride is 91.32 mmHg

We suppose the total volume = 100 mL

Step 2: Calculate volume benzene and CCl4

40 % benzene = 40 mL

60 % mL CCl4 = 60 mL

Step 3: Calculate mass benzene

Mass = density * volume

Mass of benzene = 40.00 mL *  0.87865 g/mL = 35.146 g

Step 4: Calculate moles of benzene

Moles = mass / molar mass

Number of moles of benzene  = 35.146 grams / 78 g/mol  = 0.45059 mol

Step 5: Calculate mass of CCl4

Mass of CCl4 = 60 mL * 1.5940 g/mL = 95.64 g

Step 6: Calculate moles CCl4

Number of moles of CCl4 = 95.64 grams / 154g/mol = 0.62104 mol

Step 7: Calculate total number of moles

Total number of moles = moles benzene + moles CCl4

0.45059 moles + 0.62104 moles = 1.07163 mol

Step 8: Calculate mole fraction benzene and CCl4

Mole fraction = moles benzene / total moles

Mole fraction of benzene = 0.45059 / 1.07163 = 0.4205

Mole fraction of CCl4 = 0.62104 / 1.07163 = 0.5795

Step 9: Calculate partial pressure

Partial pressure of benzene = 0.4205 * 74.61 = 31.37 mmHg

Partial pressure of CCl4      = 0.5795 * 91.32 = 52.92 mmHg

Total vapor pressure = 31.37 + 52.92 = 84.29 mmHg

The total vapor pressure is 84.29 mmHg

7 0
2 years ago
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