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2 years ago

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For the atoms that do not follow the octet rule state how many electrons surround these atoms. Express your answers as integers

separated by commas. NNON N O, NBF3, NICl2−, NOPBr3, NXeF4 = nothing electrons

1 answer:

wolverine [178]2 years ago

7 0

Answer:

The octet rule isn't all its cracked up to be. It works fine for some of the elements in the second period, but begins to come unglued for many others.

NNO does follow the octet rule, and has at least three resonance structures

N=N=O <−> N≡N−O <−> N−N≡O

Formal charges suggest that the one in the middle is most likely.

NBF3 -- Really? Show me, because I don't think there is such a compound.

NICl2^- or NICl^2- ..... which is it?

NICl2 (neutral) is an analogue of NCl3 where iodine has replaced one chlorine. It follows the octet rule.

Explanation:

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If Co(NH3)63+ has a λmax at 440 nm, calculate ΔE for the complex. A) 2.72 x 10-4 kJ/mol B) 4.52 x 10-2 kJ/mol C) 2.72 x 10 2 kJ/

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<u>Answer:</u> The energy of the complex is $2.72\times 10^2kJ$

<u>Explanation:</u>

To calculate the energy of the complex, we use the equation given by Planck which is:

$\Delta E=\frac{N_Ahc}{\lambda}$

where,

$\lambda$ = Wavelength of the complex = $440nm=4.40\times 10^{-7}m$ (Conversion factor: $1m=10^9nm$ )

h = Planck's constant = $6.624\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

$\Delta E$ = energy of the complex

Putting values in above equation, we get:

$\Delta E=\frac{6.022\times 10^{23}\times 6.624\times 10^{-34}\times 3\times 10^8}{4.40\times 10^{-7}}\\\\\Delta E=2.72\times 10^{5}J=2.72\times 10^2kJ$

Conversion factor used: 1 kJ = 1000 J

Hence, the energy of the complex is $2.72\times 10^2kJ$

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At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2

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Answer is: 0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

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Which describes any compound that has at least one element from group 17?

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The answer is HALIDE.

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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.

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