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Maru [420]
3 years ago
10

Give the following for SO2 and BrF5:

Chemistry
1 answer:
frosja888 [35]3 years ago
6 0

Answer:

The given molecules are SO2 and BrF5.

Explanation:

Consider the molecule SO2:

The central atom is S.

The number of domains on S in this molecule is three.

Domain geometry is trigonal planar.

But there is a lone pair on the central atom.

So, according to VSEPR theory,

the molecular geometry becomes bent or V-shape.

Hybridization on the central atom is

sp^{2}.

Consider the molecule BrF5:

The central atom is Br.

The number of domains on the central atom is six.

Domain geometry is octahedral.

But the central atom has a lone pair of electrons.

So, the molecular geometry becomes square pyramidal.

The hybridization of the central atom is sp^{3} d^{2}.

The shapes of SO2 and BrF5 are shown below:

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Explanation:

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Did you hear that Oxygen is going out with Magnesium?
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The ideal gas constant, R has several different values that could be used. Which quantity causes these differences? pressure tem
Blababa [14]

Answer : The correct option is, pressure.

Explanation :

The ideal gas equation is,

PV=nRT

where,

P = pressure of the gas

V = volume of the gas

n = number of moles of gas

T = temperature of the gas

R = gas constant

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R=0.08205L.atm/mol.K

R=8.3145L.kPa/mol.K

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3 0
3 years ago
Calculate the ph at the equivalence point for the titration of 0.230 m methylamine (ch3nh2) with 0.230 m hcl. the kb of methylam
Vlad1618 [11]

Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of [OH^-] be x

Initial concentration of [CH3NH_2], c = 0.230 M

        CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-

at eq'm  c-x                         x                    x

Expression of K_b:

K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}

Since ,methyl-amine is a weak base,c>>x so c-x\approx c.

K_b=\frac{x^2}{c}=5.0\times 10^{-4}=\frac{x^2}{0.230 M}

Solving for x, we get:

x=1.07\times 10^{-2} M

Given, HCl with 0.230 M , it dissociates fully in water which means [H^+] = 0.230 M

[OH^-]=[H^+] will result in neutral solution, since [OH^-]

Remaining [H^+] after neutralizing [OH^-]ions

[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M

pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65

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