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Vesna [10]
3 years ago
8

A supposedly silver crown is tested to determine its density. It displaces 10.7 ml of water and has a mass of 112 g. Part a coul

d the crown be made of silver? Could the crown be made of silver? No, the crown did not make of silver. The obtained density is higher than the density of silver which is 10.5 g/ml. Yes, the crown made of silver. The obtained density is equal to the density of silver which is 10.5 g/ml. No, the crown did not make of silver. The obtained density is lower than the density of silver which is 10.5 g/ml.
Chemistry
2 answers:
pickupchik [31]3 years ago
4 0

Answer:

As the silver crown is tested and it displaced 10.7mL of water with mass of crown  = 112g

Thus density = mass of metal / volume of water displaced = 112 / 10.7 = 10.5 g /mL

The actual density of silver = 10.5g / mL

Thus the crown is made up of silver as the obtained density is same as the actual density of silver

Vikentia [17]3 years ago
4 0

Density of a substance is the mass per unit volume of that substance. Density values of a substance are constant.

Given the mass of the crown = 112 g

Volume of water displaced by the crown = 10.7 mL

Density of the crown = \frac{Mass}{Volume}

                                   = \frac{112 g}{10.7 mL} =10.5\frac{g}{mL}

The crown is supposed to be made of silver. The density of silver is 10.5g/mL. The calculated value of density from given mass and volume displaced by the crown is equal to the theoretical density value of silver.

So the correct answer will be that the crown is made of silver as the estimated density is equal to that to density of silver.

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4 0
2 years ago
2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an
Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression d=\frac{m}{V} that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL

d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL

d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL

d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL

The problem says that all the samples have the same mass, so:

m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

V_{lithium}=\frac{m}{d_{lithium}}

V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m

V_{lithium}=1.88\frac{mL}{g}*m

V_{gold}=\frac{m}{d_{gold}}

V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m

V_{gold}=5.18*10^{-2}\frac{mL}{g}*m

V_{aluminum}=\frac{m}{d_{aluminum}}

V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m

V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m

V_{lead}=\frac{m}{d_{lead}}

V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m

V_{lead}=8.85*10^{-2}\frac{mL}{g}*m

If we assume m = 1g, we find that:

V_{lithium}=1.88mL

V_{gold}=5.18*10^{-2}mL

V_{aluminum}=3.70*10^{-1}mL

V_{lead}=8.85*10^{-2}mL

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0
3 years ago
How many grams of KF are in 2 liters of a 3.0 M solution of KF
Rufina [12.5K]

Answer:

mass ( g ) = 348 g

Explanation:

First you know : M = mole / volume (L)

in question you have the M and V and the formula of SUBSTANCE ( KF )

first you get the number of mole from equation above

so 3 = no of mole / 2

no of mole = 3 × 2 = 6 moles

and the moles equation is no of moles = mass ( g ) / molecular weight ( g/mole )

so you have already calculate the moles and you can know the MW from the Question

Mw of KF = 39 + 19 = 58

so n = mass / MW

so 6 = mass / 58

mass ( g ) = 348 g

GOOD LUCK

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3 years ago
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Mamont248 [21]

the answer is boyle's law

8 0
2 years ago
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