All categories

4 years ago

Name the 4 main layers on Earth:

Chemistry

2 answers:

ira [324]4 years ago

7 0

Answer:

s of the earth: the crust, mantle, outer core, and inner core

Explanation:

natulia [17]4 years ago

3 0

Answer:

crust,mantle,inner core, outer core

Explanation:

crust is the very outside layer, mantle is where it starts getting harder, inner core is molten rocks, and outer is a protective surface

You might be interested in

photosynthesis in plants require chloroplasts and light energy. identify two materials plants also use in this process

Nataliya [291]

I would assume carbon and water if not Im sorry.

5 0

4 years ago

A student is taking notes about the four steps in hurricane formation. Which of the following should be written as step number 1

Brrunno [24]

Tropical disturbance is the correct answer

3 0

3 years ago

In the process of preparing liquid air for fractional distillation, one fraction will be separated as a solid. What is the chemi

makvit [3.9K]

Answer:

carbon dioxide CO₂

Explanation:

Each gas has a characteristic boiling point. You can separate a random sample of gases by gradually cooling the sample until each component gas liquifies. Some compounds, such as CO₂ never liquify. Instead, they turn directly into solids.

3 0

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

How many grams of KCIO3 are needed to produce 5.00 of O2 at STP?

Arlecino [84]

2KClO₃ → 2KCl + 3O₂

mole ratio of KClO₃ to O₂ is 2 : 3

∴ if moles of O₂ = 5 mol

then moles of KClO₃ = $\frac{5 mol * 2}{3}$

= 3.33 mol

Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃

= 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol

= 407.93 g

6 0

3 years ago