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4 years ago

What two elements constitute about three fourths of earth's solid surface?

1 answer:

6 0

<span>The two elements that constitute almost three fourths of the earth's solid surface are oxygen and silicon. Oxygen accounts for 46.6 percent of the Earth's crust, while silicon is 27.7 percent. The next more plentiful element on the Earth's crust is Aluminum.</span>

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Consider a box divided in half. One side is filled with 1 mol Ne at 300. Kelvin, and the other is filled with 1 mol Rn at 300. K

vlada-n [284]

It will take the Ne gas approximately 1.2 s to equilibrate between the two sides.

We know that the time taken for a gas to diffuse is dependent on the molar mass of the gas. This is one of the interpretations of Graham's law of diffusion in gases. Thus, Let;

t1 = time taken for Rn to equilibrate = 4.0 s

t2 = time taken for Ne to equilibrate = ?

M1 = molar mass of Rn = 222g/mol

M2 = molar mass of Ne = 20 g/mol

Using

t1/t2 = √M1/M2

4/t2 = √222/20

4/t2 = 3.33

t2 = 4/3.33

t2 = 1.2 seconds

Learn more about diffusion in gases: brainly.com/question/879602

3 0

3 years ago

Calculate ΔS° for the reaction: 4Cr(s) + 3O2(g) → 2Cr2O3(s), Substance: Cr(s) O2(g) Cr2O3(s), S°(J/K⋅mol): 23.77 205.138 81.2

Mariana [72]

<u>Answer:</u> The value of $\Delta S^o$ for the reaction is 1051.93 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$4Cr(s)+3O_2(g)\rightarrow 2Cr_2O_3(s)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(Cr_2O_3(s))}=881.2J/K.mol\\\Delta S^o_{(O_2(g))}=205.13J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K$

Hence, the value of $\Delta S^o$ for the reaction is 1051.93 J/K

5 0

4 years ago

The pOH a solution of 0.04 M HCl is:________

fredd [130]

The answer is a. 1.4

Hope it helps.

6 0

3 years ago

Read 2 more answers

In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe

AnnZ [28]

The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

$\text{NH}_4\text{Cl}$ is a salt soluble in water. $\text{NH}_4\text{Cl}$ dissociates into ions completely when dissolved.

$\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq)$ .

The first test tube used to contain $\text{NH}_4\text{OH}$ . $\text{NH}_4\text{OH}$ is a weak base that dissociates partially in water.

$\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+} \;(aq)+ {\text{OH}}^{-} \; (aq)$ .

There's also an equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$ .

$\text{OH}^{-}$ ions from $\text{NH}_4\text{OH}$ will shift the equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ to the right and reduce the amount of ${\text{H}_3\text{O}}^{+}$ in the solution.

The indicator equilibrium will shift to the right to produce more ${\text{H}_3\text{O}}^{+}$ ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding $\text{NH}_4\text{Cl}$ will add to the concentration of ${\text{NH}_4}^{+}$ ions in the solution. Some of the ${\text{NH}_4}^{+}$ ions will combine with $\text{OH}^{-}$ ions to produce $\text{NH}_4\text{OH}$ .

The equilibrium between $\text{OH}^{-}$ and ${\text{H}_3\text{O}}^{+}$ ions will shift to the left to produce more of both ions.

${\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)$

The indicator equilibrium will shift to the left as the concentration of ${\text{H}_3\text{O}}^{+}$ increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

7 0

4 years ago

What did paleomagnetism give strong evidence for

ivann1987 [24]

I believe the answer is Pangaea

I hope this helps :)

3 0

3 years ago

Read 2 more answers