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3 years ago

10

PLEASE HELP!!!!

1 answer:

olga nikolaevna [1]3 years ago

4 0

8) The gravitational force is a force of attraction between two objects. It can be expressed mathematically as:

F = G m₁*m₂/d² ------(1)

m₁ and m₂ are the masses of two interacting systems

d = distance between them

G = gravitational constant

Thus according to equation(1), the gravitational force is directly proportional to the mass of the objects. Greater the masses, stronger will be the force of attraction.

In this case, the distance (d) and the one of the masses i.e. mass of earth would be the same for both situations. However, since an elephant would weigh more than a human, the gravitational force will be stronger between an elephant and earth.

9) Since the masses of spheres is the same, the gravitational force between the red and blue spheres would be the same as well, provided the distance d remains constant.

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3 years ago

In an experiment, a student needs 250.0 mL of a 0.100 M copper (II) chloride solution. A stock solution of 2.00 M copper (II) ch

LekaFEV [45]

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$M_1V_1=M_2V_2$

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$M_2\text{ and }V_2$ are the final molarity and volume of stock solution of copper (II) chloride.

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3 years ago

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