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3 years ago

2. I need to find the angle of corner of a room, what tool could I use?

Chemistry

1 answer:

Pavlova-9 [17]3 years ago

6 0

If anything you would use a protractor but that’s not a answer.... so I would pick whatever relates to a protractor

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If liquid water ___ energy it will become ice

kati45 [8]

Answer:

the answer is 1- loses

Explanation:

When water freezes it gives up some of the water's energy.

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3 years ago

What is a sun spot? in the sun on the surface

umka21 [38]

Sunspots Are decreased temperature areas. So it might be bright all around the sun spots, and the dark spots are where the temperature is lower. They usually form in polar pairs, meaning that on the other side in the same spot there would be another sunspot

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3 years ago

What is necssesary for a plant to exhibit phototropism

Solnce55 [7]

Hey The cells on the plant that are farthest from the light have a chemical called auxin that reacts when phototropism occurs.

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2 years ago

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

When solving an equation using Gay-Lussac's law, which of the following steps do not apply? A. List your knowns and unknowns. B.

suter [353]

Answer:

C. Plug the unknowns into the ideal gas law formula.

Explanation:

Gay-Lussac's law states that:

"For a constant mass of an ideal gas kept at constant volume, the pressure of the gas is directly proportional to its absolute temperature"

Mathematically, it can be written as:

$\frac{p_1}{T_1}=\frac{p_2}{T_2}$

where

p1 is the initial pressure

T1 is the initial temperature (in Kelvin)

p2 is the final pressure

T2 is the final temperature (in Kelvin)

Therefore, in order to solve for one of the 4 variables, the following steps must be adopted:

A. List your knowns and unknowns.

B. Convert temperature to kelvin.

D. Solve for the unknown.

While the following step is not needed:

C. Plug the unknowns into the ideal gas law formula.

Because we do not need to use the ideal gas law formula.

8 0

3 years ago