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4 years ago

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If the density of copper (Cu) is 8.96 g/cm3 and that of zinc (Zn) is 7.13 g/cm3. Calculate and compare the volume of a pre-1982

penny to that of a post-1982 penny? Give one reason to support your answer

1 answer:

AveGali [126]4 years ago

8 0

Answer with Explanation:

Density of copper(Cu)= $\rho_{Cu}=8.96g/cm^3$

Density of zinc(Zn)= $\rho_{Zn}=7.13 g/cm^3$

Atomic mass of copper= $63.546 g$

Atomic mass of zinc=65.38 g

We know that

Volume= $\frac{mass}{\rho}$

Using the formula

Volume of copper= $\frac{63.546}{8.96}=7.09 cm^3$

Volume of zinc= $\frac{65.38}{7.13}=9.17 cm^3$

Volume is inversely proportional to the density.

Pre -1982 penny made by copper and post-1982 penny made by zinc.

Therefore, volume of post-1982 penny is greater than the volume of pre-1982 penny because density of zinc is less than the density of copper.

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The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant,

Genrish500 [490]

Answer: (E) 300 bq

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is represented by $t_{\frac{1}{2}$

Half life of Thallium-208 = 3.053 min

Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. $\frac{2400}{2}=1200$ , after second half life, the activity would be reduced to half of 1200 i.e. $\frac{1200}{2}=600$ , and after third half life, the activity would be reduced to half of 600 i.e. $\frac{600}{2}=300$ ,

Thus the activity 9 minutes later is 300 bq.

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3 years ago

What is the total number of valence electrons un xenon(xe)

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8

Explanation:

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3 years ago

The density of lead is 11.3g/cm^3 and the density of chromium is 7.14g/cm^3. If we have one gram of each metal, what is the rati

Lady bird [3.3K]

The ratio of the volume of lead to the volume of chromium is 22 : 35

Density is simply defined as the mass of a substance per unit volume of the substance.

<h3>Density = mass / volume </h3>

To solve the question given above, we'll begin by calculating the volume of lead and chromium. This can be obtained as follow:

<h3>For Lead:</h3>

Mass of lead = 1 g

Density of lead = 11.3 g/cm³

<h3>Volume of lead =? </h3>

Density = mass /volume

11.3 = 1 / Volume

Cross multiply

11.3 × Volume = 1

Divide both side by 11.3

Volume = 1 / 11.3

<h3>Volume of lead = 0.088 cm³</h3>

<h3>For Chromium:</h3>

Mass of chromium = 1 g

Density of chromium = 7.14 g/cm³

<h3>Volume of chromium =? </h3>

Density = mass /volume

7.14 = 1 / Volume

Cross multiply

7.14 × Volume = 1

Divide both side by 7.14

Volume = 1 / 7.14

<h3>Volume of chromium = 0.14 cm³</h3>

Finally, we shall determine the ratio of the volume of lead to the volume chromium. This can be obtained as follow:

Volume of lead = 0.088 cm³

Volume of chromium = 0.14 cm³

<h3>Ratio of the volume of lead to chromium =? </h3>

Ratio = Volume of lead / Volume of chromium

Ratio = 0.088 / 0.14

Ratio = 22 / 35

<h3>Ratio = 22 : 35</h3>

Therefore, the ratio of the volume of lead to the volume chromium is 22 : 35

Learn more: brainly.com/question/17262276

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3 years ago

What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C

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Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at $26.5^oC$ = ?

$P_2$ = final pressure at $-5.1^oC$ = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $26.5^oC=273+26.5=299.5K$

$T_2$ = final temperature = $-5.1^oC=273+(-5.1)=267.9K$

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

$\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]$

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$P_1=26.5mmHg$

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