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kolezko [41]
2 years ago
14

If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the struct

ure The central X atom is single bonded to two Y atoms, which are 180 degrees apart. The X atom has two lone pairs. could be abbreviated as XY2Z2. Classify each molecule according to its shape.Molecular Geometries: (A) linear(B) bent 120(C) bent 109(D) trigonal pyramidal(E) T-shaped(F) see saw(G) square planar(H) square pyramidalPossiblities:(1) XY4Z2(2) XY5Z(3) XY2Z(4) XY2Z3(5) XY4Z(6) XY2Z2(7) XY3Z2(8) XY3Z
Chemistry
1 answer:
sashaice [31]2 years ago
5 0

Answer:

(C) bent 109

(2) XY_5Z

Explanation:

The number of bond pairs = 2

Number of lone pairs = 2

The general formula is - XY_2Z_2

According to VSEPR theory, the atoms will form a geometry in such a way that there is minimum repulsion and maximum stability.  

The central atom has 2 bond pairs and 2 lone pairs. Hybridization is sp³. The geometry is bent shape which has an angle of approximately 105.5°. It original has angle of 109.5° but due to lone pair repulsion, it reduces.

Answer - (C) bent 109

Square pyramidal possibility:- (2) XY_5Z

The central atom has 5 bond pairs and 1 lone pair. Hybridization is sp³d. The geometry is square pyramidal in which the equatorial bonds has an angle of 120° and axial bond has an angle of 90°.

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Hydrogen sulfide (H2S) is a common and troublesome pollutant in industrial wastewaters. One way to remove H2S is to treat the wa
Kryger [21]

Explanation:

As the given reaction is as follows.

     H_{2}S(aq) + Cl_{2}(aq) \rightarrow S(s) + 2H^{+}(aq) + 2Cl^{-}(aq)

So, according to the balanced equation, it can be seen that rate of formation of Cl^{-} will be twice the rate of disappearance of H_{2}S .

And, it is known that rate of disappearance of reactant will be negative and rate of formation of products will be positive value.

This means that,

Rate of the reaction = -Rate of disappearance of H_{2}S

                 = k[H_{2}S][Cl_{2}]

                 = (3.5 \times 10^{-2}) \times (2 \times 10^{-4}) \times (2.8 x 10^{-2})

                 = 1.96 \times 10^{-7} M/s

Therefore, calculate the rate of formation of Cl^{-} as follows.

Rate of formation of Cl^{-} = 2 \times 1.96 \times 10^{-7}

                                        = 3.92 \times 10^{-7} M/s

Thus, we can conclude that the rate of formation of Cl^{-} is 3.92 \times 10^{-7} M/s.

5 0
2 years ago
An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered
Karo-lina-s [1.5K]

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

The yellow flame colour indicates the presence of Na^+ which must come from the presence of Na2SO4.

5 0
1 year ago
A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav
dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of _{19}^{40}\textrm{K} radioisotope follows:

Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles  

Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles  

_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e

By the stoichiometry of above reaction,

1 mole of _{40}^{91}\textrm{Zr} is produced by 1 mole _{41}^{91}\textrm{Nb}

So, 0.17 moles of _{40}^{91}\textrm{Zr} will be produced by = \frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}

Amount of _{82}^{212}\textrm{K}

decomposed will be = 0.17 moles

Initial amount of _{40}^{91}\textrm{Nb}  will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

a=\frac{a_o}{2^n}

where,

a = amount of reactant left after n-half lives = 0.024

a_o = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

0.024=\frac{0.194}{2^n}

n=3

Therefore, 3 half lives have passed.

3 0
2 years ago
Read 2 more answers
The Lewis representation above depicts a reaction between hydrogen (blue) and a main-group element from group 5A (red). In this
Andrei [34K]

Answer:

Each Y atom needs three electrons to complete its octet by forming three bonds to hydrogen. There is one unshared pair of electrons and three bonding pairs of electrons. The bonds in the product are covalent.

Explanation:

Recall that group 5A elements contain five electrons in their outermost shell. These five electrons consists of a lone pair and three electrons that can form three bonds with hydrogen to give YH3 where Y is the group 5A element.

The YH3 molecule contains one lone(unshared) pair of electrons as well as three bonding pairs of electrons. The compounds are covalent.

3 0
2 years ago
Calculate the oxidation number of s in S2O8^2-​
mixer [17]

Given problem:

    S₂O₈²⁻

Find the oxidation number of S;

Oxidation number presents the extent of oxidation of each atom of elements a molecular formular or formula unit or an ionic radical.

  For radicals:

          "the algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion  "

 S₂O₈²⁻;   oxidation number of O is usually -2

             2(S) + 8(-2) = -2

               2S - 16  = -2

               2S = -2 + 16

                2S  = 14

                   S  = +7

The oxidation state of S in the radical is +7

3 0
2 years ago
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