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postnew [5]
3 years ago
8

I need help balancing chemical equations

Chemistry
1 answer:
Gwar [14]3 years ago
4 0
Hey! The best way I learned is if when I write it out put a space where a number should be
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As a temperature of Lava increases, ___________.
levacccp [35]
B its viscosity decreases
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3 years ago
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51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
Rolls of foil are 302 mm wide and 0.018 mm thick. (The density of foil is 2.7 g/cm3 .)
Kruka [31]

Answer:

The length of foil will be 8107.81 cm or 81.7081 m.

Explanation:

Given data:

Width of roll of foil = 302 mm

Height or thickness = 0.018 mm

Density of foil = 2.7 g/cm³

Mass of foil = 1.19 Kg

Length of foil = ?

Solution:

d = m/ v

v = length (l) × width (w) × height (h)

First of we will convert the Kg into gram and mm into cm.

one Kg = 1000 g

1.19 × 1000 = 1190 g

one cm = 10 mm

302 / 10 = 30.2 cm

0.018 / 10 =  0.0018 cm

Now we will put the values in formula:

d = m/ l× h× w

l = m / d × h× w

l = 1190 g / 2.7 g/cm³× 30.2 cm × 0.0018 cm

l = 1190 g/ 0.146772 g/cm

l = 8107.81 cm or 81.7081 m

3 0
3 years ago
The process by which hot and cold air are transferred in the atmosphere is
Elan Coil [88]

Convection: the movement caused within a fluid by the tendency of hotter and therefore less dense material to rise, and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat.

hope that helps :)

3 0
3 years ago
¿Cuántos lunares hay en 1,035 gramos plomo <br>​
SpyIntel [72]

Answer:

TERO BAU KO CHAK

...BUJHIS

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3 years ago
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