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Amiraneli [1.4K]
2 years ago
7

The density of mercury (the element) is 13.69 g/cm3. The density of Mercury (the planet) is 5.43 g/cm3. If Mercury (the planet)

was made of mercury (the element) and had the same mass, what would its diameter be?
Chemistry
1 answer:
Annette [7]2 years ago
5 0

Answer:

The diameter of Mercury would be 0.735 times smaller than the actual diameter i.e., it would become 3590 km from 4879 km

Explanation:

For the sake of simplicity, let us consider Mercury spherical and uniform. The provided data shows that if the Mercury planet is purely composed of mercury then it will require less space as compared to the actual planet, due to the higher density of the mercury. The ratio of the two densities shows that the volume of Mercury would reduce 2.52 times. From the change in volume we can determine the change in diameter by taking the ratio of initial and final states, shown in the following derivation:

V=(\frac{4}{3})(\pi )(r^{3}),

Here, V is the volume and r is the radius of the Mercury.

From the ratio of initial and final states, we obtain the following relation

\frac{V_{2} }{V_{1} } = \frac{r_{2} ^{3}}{r_{1} ^{3}}

as,

V_{2}=\frac{V_{1}}{2.52}

so,

\frac{1}{2.52} = \frac{r_{2} ^{3}}{r_{1} ^{3}}

r_{2}^{3}=\frac{r_{1}^{3}}{2.52}

r_{2}=  0.735.r_{1}

The factor will remain same for the diameter i.e.,

D_{2}= 0.735 D_{1}

For absolute diameter of Mercury:

Actual diameter of Mercury = 4879.4 km

Hence, the diameter of Mercury, if it was made of mercury would become 3590 km.

P.S.: The element mercury first letter is kept small while the other is capped

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How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
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<span>Answer: 100 ml
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<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
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iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
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<span>2) Use the percent yield to calculate the theoretical amount:
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<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
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<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
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<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
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=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
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<span>i) Molarity formula: M = n / V
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<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
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Answer:

None of these

Explanation:

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Step -1 : Generation of stable carbocation.

Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.

Step-2: Attack of the ring to the carbocation

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