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baherus [9]
3 years ago
12

A graduated cylinder with just water has a volume of 20.0 mL. When a 25.0 g piece of zinc alloy is added, the new volume is 23.2

mL. What is the density of the zinc alloy? (write your answer to 2 sig figs)
Chemistry
1 answer:
Dmitry_Shevchenko [17]3 years ago
7 0

Density of any substance is mass of the substance per unit volume. Density, d=m/V, where m is mass of the substance and V is the volume occupied by the substance. The amount of water displaced by an object when it is  submerged in water equals the volume of the object. So, due to submerge of piece of zinc alloy in water  present in graduated cylinder, water level rises. The increase in the volume of water is the volume of the zinc piece.

Volume of water in graduated cylinder before addition of zinc piece (v₁)= 20 ml, Volume of water in graduated cylinder after addition of zinc piece (v₂)= 23.2 ml, Rise in the volume due to submerge of the piece = (v₂ - v₁)ml= 23.2-20.0= 3.20 ml. So, volume of the piece is 3.20 ml and mass of the zinc piece is  25.0 g.

Hence density zinc alloy is equal to \frac{25.0}{3.20} g/ml= 7.81 g/ml= 7.8 g/ml (expressed in two significant figure).


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Answer:

<em>Dentro de las aplicaciones de la óxido-reducción se pueden encontrar:</em>

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A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam
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Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

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