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notka56 [123]
3 years ago
14

Which of these molecules has the same number of shared electron pairs as unshared electron pairs? (a) hcl, (b) h2s, (c) pf3, (d)

ccl2f2 (e) br2?
Chemistry
2 answers:
Paul [167]3 years ago
8 0
<span>H2S has 2 bonds which means that it has two shared electron pairs. It has 2 lone pairs on electrons. It's the only one where the 2 are the same.</span>
uranmaximum [27]3 years ago
7 0

Let us check each molecule

a) HCl : the number of shared electron pairs is "one"

the number of unshared electron pairs is six [on chlorine]

b) H₂S : the number of shared electron pairs is "two" as there are two bonds of sulphur with two hydrogen atoms.

The number of valence electrons on sulphur are six, out of these two are involved in bonding and rest four are in the form of two unshared (lone pair) pair of electrons

So in H₂S it has the same number of shared electron pairs as unshared electron pairs

c) PF₃ : the shared pair of electrons are "three"

the number of unshared electron pairs is one on phosphorous.

d) CCl₂F₂ : there are total four shared pair of electrons on carbon and zero unshared electron pairs

e) Br₂ : only one shared electron pair and three unshared electron pairs

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Given a fixed amount of gas help at a constant pressure, calculate the temperature to which the gas would have to be changed if
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592 K or 319° C

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From the statement of Charles law we know that the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure. Thus;

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What mass of ice can be melted with the same quantity of heat as required to raise the temperature of 3.00 mol H2O(l) by 50.0°C?
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Answer:

m=33.9g

Explanation:

Hello,

In this case, we can first compute the heat required for such temperature increase, considering the molar heat capacity of water (75.38 J/mol°C):

Q=nCp \Delta T=3.00mol*75.38\frac{J}{mol\°C} *50.0\°C\\\\Q=11307J

Afterwards, the mass of ice that can be melted is computed by:

Q=n \Delta _{fus}H

So we solve for moles with the proper units handling:

n=\frac{Q}{\Delta _{fus}H} =\frac{11307J}{6010\frac{J}{mol} } =1.88mol

Finally, with the molar mass of water we compute the mass:

m=1.88mol*\frac{18g}{1mol}\\ \\m=33.9g

Best regards.

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