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Sergeeva-Olga [200]
3 years ago
8

A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme

rged in the graduated cylinder, the total volume increases to 61.2 ml. determine the density of this piece of jewelry. assuming that the jewelry is made from only one substance, what substance is it likely to be? explain.
Chemistry
1 answer:
Simora [160]3 years ago
4 0

Answer:- Density is 10.5 grams per mL and the substance is Silver(Ag).

Solution:-  Mass of jewelry is given as 132.6 g. Initial volume of water in the graduated cylinder is 48.6 mL and the volume becomes 61.2 mL when this jewelry is submerged in the cylinder.

So, the volume of the jewelry = 61.2 mL - 48.6mL = 12.6 mL

We know that, density=\frac{mass}{volume}

Let's plug in the values in the above formula to calculate the density:

density=\frac{132.6g}{12.6mL}

density=10.5\frac{g}{mL}

So, the density of the jewelry is 10.5\frac{g}{mL} .

The substance from which the jewelry is made could easily be identified by comparing the calculated density with the density values given for different substances in the density table.

Looking at the density table, 10.5 grams per mL is the density for silver. So, the jewelry is made up from Silver.

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What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of
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Answer:

25.76 L

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Considering the expression for density as:

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So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

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Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{31.862\ g}{63.546\ g/mol}

<u>Moles of copper = 0.5014 moles </u>

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

Density=\frac {Mass}{Volume}

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So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,  

Mass of nitric acid = \frac {68}{100}\times 284\ g = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

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Moles= \frac{193.12\ g}{63.01\ g/mol}

<u>Moles of nitric acid = 3.0649 moles </u>

According to the reaction,  

Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}

1 mole of copper react with 4 moles of nitric acid

Thus,  

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

<u>Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent. </u>

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

<u>Moles of nitrogen dioxide = 1.0028 moles </u>

Given:  

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P(torr)=\frac {1}{760}\times P(atm)

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Pressure = 735 / 760 atm = 0.9632 atm

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The conversion of T( °C) to T(K) is shown below:

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T₁ = (28.2 + 273.15) K = 301.35 K  

Using ideal gas equation as:

PV=nRT

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P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

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0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K  

<u>⇒V = 25.76 L</u>

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