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kogti [31]
3 years ago
14

Based on the Valence Shell Electron Pair Repulsion Theory (or VSEPR), molecules will arrange to keep the following as far apart

as physically possible
a) mobile electrons
b) valence electron pairs
c) inner shell electrons
d) the electrons closest to the nuclei
Chemistry
1 answer:
elena55 [62]3 years ago
6 0

Answer:

B. Valence Electron Pairs

Explanation:

Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.

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In addition to those in Table 14.3, other less stable nitrogen oxides exist. Draw a Lewis structure for each of the following:(c
Natalka [10]

Lewis structure for each of the following N₂O₃ with no N¬N bond is attached below.

Even though pi symmetry occupies the antibonding orbitals of NO, this is unimportant after the dimer forms. A sigma connection exists. The enthalpy of the newly formed sigma bond in the dimer is low because the loss of a particularly distinctive set of single-electron resonance forms that were available for no monomer offset the net gain in bond. When the whole free energy is taken into account, there is no gain because the entropic effects are on the order of 1030kJ/mol, and dimerization is entropically disfavored at G=17kJ/mol. Therefore, any little increase in enthalpy is cancelled out by the loss of entropy.

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5 0
2 years ago
Why is temperature not a chemical change
GrogVix [38]

Answer:

Temperature is not a chemical change because when a substance changes in temperature, its chemical makeup is not changing.

6 0
3 years ago
Read 2 more answers
. The Henry's law constant for helium gas in water at 30 ∘C is 3.7×10−4M/atm; the constant for N2 at 30 ∘C is 6.0×10−4M/atm. a.
ludmilkaskok [199]

Explanation:

1) Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

M_{g}=K_H\times p_{g}

where,

K_H = Henry's constant

p_{He} = partial pressure of gas

a)  K_H=3.7\times 10^{-4} M/atm

p_{He}=2.1 atm

Putting values in above equation, we get:

M_{He}=3.7\times 10^{-4} M/atm\times 2.1 atm = 7.77\times 10^{-4} M

The solubility of helium gas is 7.77\times 10^{-4} M

b)  K_H=6.0\times 10^{-4} M/atm

p_{N_2}=2.1 atm

Putting values in above equation, we get:

M_{He}=6.0\times 10^{-4} M/atm\times 2.1 atm = 1.26\times 10^{-3} M

The solubility of nitrogen gas is 1.26\times 10^{-3} M

2)

a) Mass of solute or methanol , m= 14.7 g

Mass of solvent or water , m'= 186 g

Mass of the solution = M = m + m' = 14.7 g + 186 g = 200.7 g

w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100

=\frac{14.7 g}{200.7 g}\times 100=7.32\%

The mass percent of methanol is 7.32%.

b) Molality is defined as moles of solute per kilograms of solvent.

m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}

Moles of methanol = \frac{14.7 g}{32 g/mol}=0.4594 mol

Mass of solvent = 186 g = 0.186 kg

m=\frac{0.4594 mol}{0.186 kg}=2.4610 mol/kg

The molality of methanol is 2.4610 mol/kg.

4 0
3 years ago
Why am i so cool<br><br> insta: gabriel.the.businessman
mario62 [17]

Answer:

1 +1 = 2

(Just put this here so it doesn't get flagged)

Explanation:

1 +1 = 2

2 -1 =1

1-1 =0

Ta da

Q: Why am I so cool?

A: I have no clue lol :)

Could you please mark this answer as brainiest?

It would help me a lot!

Thank you!

5 0
3 years ago
How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

Thus,

\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

t = 6\times t_{1/2}

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0
4 years ago
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