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3 years ago

13

A balloon occupies 175 cm at 25.0 C. What volume will it occupy at 42.5 °C?

1 answer:

patriot [66]3 years ago

4 0

175 cm³ instead of 175 cm.

Answer:

297.5 cm³

Explanation:

From the question;

Initial volume; V1 = 175 cm³

Initial temperature; T1 = 25°C

Final temperature; T2 = 42.5°C

From Charles law we can find the volume V2 from the equation;

V1/T1 = V2/T2

Making V2 the subject gives;

V2 = V1 × T2/T1

V2 = (175 × 42.5)/25

V2 = 297.5 cm³

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Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu

IrinaK [193]

Answer : The atomic radius for Ti is, $1.45\times 10^{-8}cm$

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

First we have to calculate the volume of HCP crystal structure.

Formula used :

$\rho=\frac{Z\times M}{N_{A}\times V}$ .............(1)

where,

$\rho$ = density = $4.51g/cm^3$

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass = 47.87 g/mole

$(N_{A})$ = Avogadro's number

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

$4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}$

$V=1.06\times 10^{-22}cm^3$

Now we have to calculate the atomic radius for Ti.

Formula used :

$V=6R^2c\sqrt{3}$

Given:

c/a ratio = 1.669 that means, c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

$V=6R^2\times (1.669a)\sqrt{3}$

$V=6R^2\times (1.669\times 2R)\sqrt{3}$

$V=(1.669)\times (12\sqrt{3})R^3$

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

$1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3$

$R=1.45\times 10^{-8}cm$

Therefore, the atomic radius for Ti is, $1.45\times 10^{-8}cm$

3 0

3 years ago

A solute that has been dissolved in a solvent ________.

grigory [225]

Answer:

a solution: for example when sugar is dissolved in water it becomes a sugar solution

8 0

2 years ago

A sample consisting of 2 moles He is expanded isothermally at 0 degrees from 5.0dm3 to 20.0dm3. Calculate w, q and deltaU for ea

FinnZ [79.3K]

Answer:

i) $\Delta U=0 w=-6293 J q=6293 J$

ii) $\Delta U=0 w=-3404,52 J q=3404,52 J$

ii) $\Delta U=0 w=0 J q=0 J$

Explanation:

As the initial and final states of the sample are the same, the ΔU of the sample is, for the three cases

$\Delta U=n.C_{V}.\Delta T=0$ since $\Delta T=0$

i)Reversibly $P_{ext} =P_{sys}$ so $w$ can be calculated by

$w=-n.R.T.ln(\frac{V_{f}}{V_{i}})=-2 \times 8.314\frac{J}{mol K} \times 273,15K \times ln(\frac{}{5dm^{3}})=-6293 J$

and because of the first law of thermodynamics

$q=-w=6293 J$

ii)Irreversibly with $P_{ext} =P_{f}$

we can calculate $P_{f}$ by the law of ideal gases

$P_{f} =\frac{n\times R\times T}{V_{f}} =\frac{2\times 0.082\frac{dm^{3}atm}{mol K}\times 273,15K}{20dm^{3}} =2,24 atm$

then w can be calculated by

$w=-P_{ext} \times \Delta V=-2,24 atm \times (20-5) dm^{3} \times frac{101.325J }{atm dm^{3}=-3404,52J$

and

$q=-w=3404,52J$

iii)a free expansion $P_{ext}$ so $w=0$ (there's no work at vaccum) and $q=-w=0$

7 0

3 years ago

Which is a correct comparison between the modern quantum model and John Dalton’s model of the atom?

Greeley [361]

Answer:

The modern model recognizes particles in the atom, whereas Dalton’s model does not.

Explanation:

The modern model considers the atom to consist of electrons, protons, and neutrons.

A is wrong. The modern model of the atom is widely accepted.

B is wrong. Dalton assumed that all atoms are identical. The modern model recognizes that there are isotopes of atoms.

D is wrong. The modern model explains emission spectra. Dalton's model does not.

4 0

4 years ago

Read 2 more answers

I need to know how to find atoms on a periodic table.

Juli2301 [7.4K]

Answer:

you get atoms number left down the element example hydrogen has 1 atoms number and mass number.

7 0

2 years ago