Chemistry, 19.05.2020 20:09 emmamood

Consider the molecule below. Determine the hybridization at each of the 2 labeled carbons. There is HF with Cl single-bonded to the second (from left to right) carbon and CH3 single-bonded to the fourth carbon. The first and the third carbons both have H single-bonded. There is a double bond between the first and the second and between the third and the fourth carbons. The second carbon of the chain is labeled as 1 and the carbon of the CH3 group is labeled as 2. Consider the molecule below. Determine the hybridization at each of the 2 labeled carbons. There is HF with Cl single-bonded to the second (from left to right) carbon and CH3 single-bonded to the fourth carbon. The first and the third carbons both have H single-bonded. There is a double bond between the first and the second and between the third and the fourth carbons. The second carbon of the chain is labeled as 1 and the carbon of the CH3 group is labeled as 2. C1 = sp2, C2 = sp3d C1 = sp2, C2 = sp2 C1 = sp3d, C2 = sp3d2 C1 = sp3, C2 = sp2 C1 = sp2, C2 = sp3

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Answer from: Quest

Ibelieve it's celsius. hope this .

Answer from: Quest

So3 that’s what i think

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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Consider the molecule below. Determine the hybridization at each of the 2 labeled carbons. There is...

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