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3 years ago

Which statement is true about the electrons that can be located together in an orbital?

Chemistry

1 answer:

Salsk061 [2.6K]3 years ago

3 0

The two electrons that share an orbital repel each other.

All electrons bear a negative charge. They are held in their orbits by the attractive force of charged protons. The farther away an orbital is to the atomic nucleus the easier it is to expunge an electron from this distant orbital shell.

Explanation:

Because electrons have the same negative charge, they repel each other especially when they occupy the same orbital shell in an atom. To reduce this repulsion, each of the electrons in the orbital shell (remember electrons occupy orbital shells of atoms in 2s) assumes an opposite quantum (Ms) spin; one with – ½ while the other + ½ .

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Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

Answer: The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

Explanation:

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

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

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

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

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

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

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

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

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

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

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

7 0

3 years ago

Sam toured India with his class. He saw many buildings that reminded him of buildings he saw in Saudi Arabia and in London.

melomori [17]

A) cultural diffusion

4 0

2 years ago

Read 2 more answers

Uranium-238 decays to lead-206 with a half-life of 4.5 x 109 yr. Determine how much uranium-238 decays in milligrams (to three s

Mamont248 [21]

Answer:

2.15 mg of uranium-238 decays

Explanation:

For decay of radioactive nuclide-

$N=N_{0}.(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$

where N is amount of radioactive nuclide after t time, $N_{0}$ is initial amount of radioactive nuclide and $t_{\frac{1}{2}}$ is half life of radioactive nuclide

Here $N_{0}=4.60 mg$ , $t=4.1\times 10^{9}yr$ and $t_{\frac{1}{2}}=4.5\times 10^{9}yr$

So, $N=(4.60mg)\times (\frac{1}{2})^{\frac{4.1\times 10^{9}}{4.5\times 10^{9}}}$

so, N = 2.446 mg

mass of uranium-238 decays = (4.60-2.446) mg = 2.15 mg

3 0

3 years ago

Listed below are possible solutes. In your possession is a beaker with 500 milliliters of water. You want to make an aqueous sol

slavikrds [6]

Answer: I & III

Explanation: Solutes are the substances which are minimum in quantity and which is required to dissolve in the solvent (which is larger in quantity) in order to make a solution.

In the asked question, it is given that the water is the solvent and from the given solutes we have to pick which would make an aqueous solution with the highest concentration of solute possible.

Thus the most appropriate answers could be the Ammonia and hexanol which can make the highest possible concentration of solute as ammonia is the gas which is highly soluble in water and hexanol is an alcohol which has an affinity for water. Thus the correct option is I & III

8 0

3 years ago

Read 2 more answers

calculate the ph for each case in the titration of 50.0 ml of 0.130 m hclo(aq) with 0.130 m koh(aq). use the ionization constant

-BARSIC- [3]

The ph before the addition of any Koh is 10.105.

Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.

After the addition of 50 ml KOH,

moles of KOH = 50 * 0.13 = 6.5 mmol

moles of HClO = 50 * 0.13 = 6.5 mmol

occurred hydrolysis solution,

pH = 0.5(14 + pKa + log [base conjugate])

pH = 0.5(14 + (- log (4 * 10^-8)) + log (6.5/(50 + 50)))

pH = 10.105

The concentration of a substance is the quantity of solute found in a given amount of solution. Concentrations are normally expressed in terms of molarity, defined because of the variety of moles of solute in 1 L of answer.

The Concentration of an answer is a measure of the quantity of solute that has been dissolved in a given amount of solvent or answer. A concentrated answer is one that has a rather huge quantity of dissolved solute.

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4 0

1 year ago