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pshichka [43]
3 years ago
14

The hybrid orbital set used by the central atom in h2co is ________. the hybrid orbital set used by the central atom in h2co is

________. sp sp2 sp3 sp3d sp3d2

Chemistry
1 answer:
Feliz [49]3 years ago
5 0
The Lewis structure for H₂CO is shown in the attached picture. The central atom is the carbon. However, I'm not sure which bond you're referring to. There can be two answers. The two C-H bonds are sp³ hybridized because it is a single bond. The C=O bond is sp² hybridized because it is a double bond.

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Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol[/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0
3 years ago
Determine the mass of 2.75 moles of CaSO4. Record your work and your answer.
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Explanation:

RFM \:  = 160 \: g \\ 1 \: mole \: weighs \: 160 \: g \\ 2.75 \: moles \: weighs \:  \frac{2.75 \times 160}{1} g \\  = 440 \: g

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Suppose that a dependent quantity increases when the independent quantity is decreased. This kind of proportional relationship b
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I thinks it’s not this article we should cajnges it
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3 years ago
A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s
DerKrebs [107]

Answer:

11.31g NaClO₂

Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>

You can change the concentration of the substance if you write the moles of the substances:

[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>

Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =

<h3>11.31g NaClO₂</h3>
5 0
3 years ago
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