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Mars2501 [29]
3 years ago
9

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

four substances whose properties are given, which is the most likely identity of this substance? Substance Density (g/cm3) Melting Point (°C) Boiling Point (°C) molybdenum 10.28 2,623 4,639 mercury 13.53 -39 357 hafnium 13.31 2,233 4,603 lead 11.34 327 1,749 A. molybdenum B. mercury C. hafnium D. lead
Chemistry
2 answers:
Tpy6a [65]3 years ago
6 0

Answer:

It is (C) Hafnium

Explanation:

horrorfan [7]3 years ago
3 0

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution :  Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 Cm^{3}

First, find the Density of given substance.

Formula used :    

Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}

Now,put all the values in this formula, we get

Density=\frac{46.9 g}{3.5 Cm^{3} } = 13.4 g/Cm^{3}

So, we conclude that the density of given substance (13.4 g/Cm^{3}) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/Cm^{3} respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

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V ( H2SO4) = 35 mL / 1000 => 0.035 L

M ( H2SO4) = ?

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n = 0.025 x 0.320 => 0.008 moles of NaOH

Mole ratio:

<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
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Therefore:

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hope this helps!



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