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Licemer1 [7]
3 years ago
6

The electron-domain geometry of a sulfur-centered compound is trigonal bipyramidal. The hybridization of the central sulfur atom

is ________.
Chemistry
1 answer:
VMariaS [17]3 years ago
8 0

The hybridization state of the central sulfur atom is sp3d.

The valence shell electron pair repulsion theory describes the shape of molecules in terms of the number of electron pairs that surround the valence shell of the central atom in the molecule.

There are certain molecular shapes that are associated with certain hybridization states of the central atom. Since the central sulfur atom is trigonal bipyramidal, the hybridization state is sp3d.

Learn more: brainly.com/question/13978941

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For the problem: Mg (s) + O₂ (g) ➞ MgO (g) if 4.3 g of magnesium reacts with 83 g of oxygen (O₂), which is the limiting reagent?
Anit [1.1K]

Explanation:

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3 0
3 years ago
Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.

7 0
3 years ago
Based on the equation 3 Cu (s) + 8 HNO3 (aq) 3 Cu(NO3)2 (aq) + 2NO (g) + 4H2O (g) how many grams of Cu would be needed to react
eimsori [14]

Answer:

Mass = 381.28 g

Explanation:

Given data:

Number of moles of HNO₃ = 16 mol

Mass of Cu needed to react with 16 mol of HNO₃ = ?

Solution:

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3Cu + 8HNO₃    →     3Cu(NO₃)₂ + 4H₂O + 2NO

Now we will compare the moles of Cu with HNO₃ from balance chemical equation.

                     HNO₃         :          Cu

                        8              :         3

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Mass of Cu needed:

Mass = number of moles × molar mass

Mass = 6 mol × 63.546 g/mol

Mass = 381.28 g

4 0
4 years ago
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lesantik [10]

Answer:

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7 0
3 years ago
Seawater is around a 3% aqueous salt solution.
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Answer:

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