In the ground state, which atom has a completely filled valence electron shell?

If 9.5 atms of pressure were increased to 25 atms of pressure, what would be the final volume

nordsb [41]

Answer:

The final volume of the gas is 36.1 L.

Explanation:

Given:

Initial pressure of the gas is, $P_1=9.5\ atm$

Final pressure of the gas is, $P_2=25\ atm$

Initial volume of the gas is, $V_1=95\ L$

Final volume of the gas is, $V_2=?$

Here, we shall use Boyle's Law which states that for a process under constant temperature, the pressure of the gas changes inversely with the change in volume.

Here, the pressure is increased. So, the volume of the gas is decreased.

Therefore, as per Boyle's Law:

$P_1V_1=P_2V_2\\9.5\times 95=25V_2\\902.5=25V_2\\V_2=\frac{902.5}{25}\\V_2=36.1\ L$

So, the final volume of the gas is 36.1 L.

4 0

3 years ago

Why did the author include details about the cooking show he saw?

Stella [2.4K]

Answer:

he wants to explain why he started cooking

6 0

3 years ago

Read 2 more answers

Quick Chemistry question, please show work :) 3. If 25.0 mL of a 4.525 M solution of HCl was poured into 112.0 mL of distilled w

fomenos

The total mole number of solutes will not change. So it is 25 * 10^-3 *4.525 mole. And the final volume is 25+112 = 137 mL. So the resulting concentration is mole number / volume=0.8257 M.

7 0

3 years ago

Choose ALL the answers that apply.

zaharov [31]

Answer: absorbs food, breaks down food

5 0

3 years ago

"Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite react

frosja888 [35]

<u>Answer:</u> The given reaction is non-spontaneous in nature.

<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:

$3Mg(s)+Cr_2O_3(s)\rightarrow 3MgO(s)+2Cr(s)$

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

$\Delta S^o_{rxn}=[(3\times \Delta S^o_{(MgO(s))})+(2\times \Delta S^o_{(Cr(s))})]-[(3\times \Delta S^o_{(Mg(s))})+(1\times \Delta S^o_{(Cr_2O_3(s))})]$

We are given:

$\Delta S^o_{(Mg(s))}=32.68J/K.mol\\\Delta S^o_{(Cr_2O_3(s))}=81.2J/K.mol\\\Delta S^o_{(MgO(s))}=26.94J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(3\times (26.94))+(2\times (23.77))]-[(3\times (32.68))+(1\times (81.2))]\\\\\Delta S^o_{rxn}=-50.88J/K=-0.0509kJ/K.mol$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

To calculate the standard Gibbs free energy of the reaction, we use the equation:

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy change of the reaction = 665.1 kJ/mol

T = Temperature = 298.15 K

$\Delta S^o$ = standard entropy change of the reaction = -0.0509 kJ/K.mol

Putting values in above equation, we get:

$\Delta G^o=(665.1kJ/mol)-(298.15K\times (-0.0509kJ/K.mol))=680.27kJ/mol$

As, the Gibbs free energy of the reaction is coming out to be positive, the reaction is non-spontaneous in nature.

Hence, the given reaction is non-spontaneous in nature.

4 0

4 years ago