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2 years ago

11

a gas that exerts a pressure of 6.20 atm in a container with a volume of ____ mL will exert a pressure of 9.150 atm when transfe

rred to a container with volume of 322 mL

2 answers:

leonid [27]2 years ago

7 0

The answer is 475 mL

Aloiza [94]2 years ago

7 0

475

6.2 / 9.15 = x / 322

Cross multiply

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

IRINA_888 [86]

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

$A=6R^{2}\sqrt{3}$

Here, R is radius and is related to a as follows:

$R=\frac{a}{2}$

Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Now, volume can be calculated as follows:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

Here, A is atomic mass of $Cr_{2}O_{3}$ is 151.99 g/mol.

Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

6 0

2 years ago

Read 2 more answers

A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?

Tanzania [10]

<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume = 0.50 cm × 10.0 cm × 2.0 cm

= 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

= 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

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2 years ago

5. Before vehicle emissions were well-regulated CO emissions were 66 g/mile. Assume this emission rate applies for an airshed. T

Aleks [24]

Answer:CALINE4 Model and Geographical Information System were used for

the present study to predict the CO concentrations and prepare thematic

maps for the study area.

CALINE4 is a latest model that predicts the concentration of carbon

monoxide impacts near the roadways. The California Department of

Transformation (CALTRANS) developed the model and its purpose is to help

planners protect public health from the adverse effects of carbon monoxide.

The model predictions along with the aid of GIS based model help to arrive at

air shed levels of carbon monoxide. CALINE4 is a simple line source

Gaussian plume dispersion model. The user defines the proposed roadway

geometry, worst-case meteorological parameters, anticipated traffic volumes

and receptor positions to predict the concentrations of pollutant.

CALINE4 is graphical with windows based user interface, designed to

ease data entry and increase the online help capabilities. The model was

developed for predicting the concentrations of relatively inert pollutants such

as carbon monoxide and it is now used for several other pollutants like NO2

and SPM. The model is based on fine tuned Gaussian diffusion equation and

it employs mixing zone concept to characterize the pollutant dispersion over

the roadways. Given the exhaust emission concentrations, meteorology and

site geometry, the model can predict the concentration of pollutants for any experimentation

Explanation:

4 0

2 years ago

How many moles of gas Does it take to occupy 520 mL at a pressure of 400 torr and a temperature of 340 k

Ann [662]

Answer would be B. I provided work on an image attached. Message me if u have any other questions on how to do it

6 0

2 years ago

The nuclear equation is incomplete. Superscript 239 Subscript 94 Baseline P u + Superscript 1 Subscript 0 Baseline n yields Supe

Likurg_2 [28]

Answer:

The correct option is the first option

Explanation:

The equation described in the question is shown below

²³⁹₉₄Pu +¹₀n ⇒ ¹⁰⁰₄₀Zr + blank + 2¹₀n

This equation is a nuclear fission because it involves the splitting of a heavy nucleus, Plutonium (Pu), into smaller nuclei, Zirconium (Zr) and an unknown nuclei.

The law of conservation of matter states that matter can neither be created nor destroyed hence in other to get the missing atom, we must know the total number of subscripts (mass number) and superscripts (atomic number) on both sides.

The total mass number on the reactant side is 239 + 1 = 240

The total atomic number on the reactant side is 94 + 0 = 94

While, The total mass number on the product side is 100 + 2(1) = 102

The total atomic number on the product side is 40 + 2(0) = 40

To determine <u>the missing atom</u>, you subtract the the total atomic number of the product from that on the reactant side; 94 - 40 = 54

The <u>atom with the atomic number 54 is Xenon</u>,

To <u>determine the mass number</u> of the xenon isotope involved, you subtract the the total mass number of the product from that on the reactant side; 240 - 102 = 138.

Hence, the particle that completes the equation is ¹³⁸₅₄Xe, the first option

5 0

2 years ago