What is the periodic table a list of

Good evening! Does anyone know this? Earth Science

riadik2000 [5.3K]

The fraction of Earth's radius (6371 km) relative to the thickness of the oceanic (7.5 km) and continental crust (35 km) is 0.12 and 0.55, respectively.

What we know:

The average radius of Earth (E) = 6371 km

The average thickness of oceanic crust (O) = 7.5 km

The average thickness of continental crust (C) = 35 km

We need to convert all the above units from kilometers to miles:

$E = \frac{0.6214 mi}{1 km}*6371 km = 3958.9 mi$

$O = \frac{0.6214 mi}{1 km}*7.5 km = 4.7 mi$

$C = \frac{0.6214 mi}{1 km}*35 km = 21.7 mi$

Now, we can calculate the fraction of Earth's radius relative to each type of crust, with the given equation:

$X = \frac{avg. \: thickness}{avg. \: radius} \times 100$

<u>For the oceanic crust (O)</u>:

$X = \frac{4.7 mi}{3958.9 mi}\times 100 = 0.12$

<u>For the continental crust (C)</u>:

$X = \frac{21.7 mi}{3958.9 mi}\times 100 = 0.55$

Therefore, the fraction of Earth's radius relative to the oceanic and continental crust is 0.12 and 0.55, respectively.

You can see another example of calculation of fractions of Earth's radius here: brainly.com/question/4675868?referrer=searchResults

I hope it helps you!

6 0

2 years ago

3. How many molecules are in 0.500 moles of sulfur?

satela [25.4K]

Explanation:

No of molecules=0.500×6.023×10²³=3.011×10²³ molecules

7 0

3 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

$PCl_3,p_1'=6.798 Torr$

$Cl_2,p_2'=26.398 Torr$

$PCl_5,p_3'=223.402 Torr$

Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=p_1=13.2 Torr$

Partial pressure of the $Cl_2=p_2=13.2 Torr$

Partial pressure of the $PCl_5=p_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{p_1}{p_1\times p_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=p_1'=13.2 Torr$

Partial pressure of the $Cl_2=p_2'=?$

Partial pressure of the $PCl_5=p_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=p_1'+p_2'+p_3'$

$263.0Torr=13.2 Torr+p_2'+217.0 Torr$

$p_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{p_3'}{p_1'\times p_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

$p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

$p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

$p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

6 0

3 years ago

For a particular isomer of C8H18, the combustion reaction produces 5104.1 kJ of heat per mole of C8H18(g) consumed, under standa

Vladimir79 [104]

Answer:

Explanation:

Hello,

Considering the chemical reaction, the enthalpy of reaction is given by:

ΔH°rxn=ΔfHCO2+ΔfHH2O-ΔfHC8H18

(ΔfHO2=0)

Taking into account that the reaction produces energy, ΔH°rxn is negative. No, solving for ΔfHC8H18:

ΔfHC8H18=-ΔH°rxn+8*ΔfHCO2+9*ΔfHH2O

ΔfHC8H18=-(-5104.1 kJ/mol)+9*(-292.74kJ/mol)+8*(-393.5 kJ/mol)

ΔfHC8H18=-678.56 kJ/mol

Best regards.

3 0

3 years ago

HELP!! what law of thermodynamics does the nuclear power plant have?

Alex777 [14]

Answer:

The amount of electric power produced for each unit of thermal power gives the plant its thermal efficiency, and due to the second law of thermodynamics there is an upper limit to how efficient these plants can be.

7 0

3 years ago