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3 years ago

Tell me the main parts of the Earth Layers.

Chemistry

1 answer:

sukhopar [10]3 years ago

8 0

Answer:

Crust (the surface)

Upper Mantle (dense magma)

Lower Mantle (less dense magma)

Outer Core (pure liquid material)

Inner Core (dense material)

Hope that helps

You might be interested in

What is one property that is different between water and oil? *

Arlecino [84]

Oil is less dense than water, so the difference would be its density. Water is a good solvent, which means It can dissolve other substances.

3 0

3 years ago

1. Mg2+ and Ca2+ are in the same group on the periodic table. In terms of electronic

finlep [7]

Answer:

Se detailed explanation.

Explanation:

Hello,

In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:

- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.

- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.

- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.

- Their electron configurations are:

$Ca^{20}=1s^2,2s^2,2p^6,3s^3,3p^6,4s^2\\\\Mg^{12}=1s^2,2s^2,2p^6,3s^2$

It means that the both of them are at the $s$ region since it is the last subshell at which their electrons are.

Best regards.

8 0

4 years ago

What is the main purpose of using NaOH in the saponification reaction . NaOH works as the catalyst b. NaOH solution is the solve

Alex777 [14]

Answer:

C. NaOH acts as a reactant in the reaction

Explanation:

Because during the saponification process, Na+ replaces the H+ in the fatty acid been used for the saponification process

7 0

4 years ago

Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

3 years ago

Please help it’s due today I will give brainliest

Thepotemich [5.8K]

1st and 4th options are suitable answers, as these 2 changes are not exactly physical changes as it cant return back to original form and as well as its not cooling, so I feel its 1st and 4th options

7 0

3 years ago