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Furkat [3]
3 years ago
9

S2- has the same electron configuration as which neutral element?

Chemistry
1 answer:
Llana [10]3 years ago
6 0

Answer:

ANSWER OS HERE

Explanation:

THE O2- has two more extra electrons due to presence of 2 negative charges so a total of 10 electrons and the element that has 10 electrons in the periodic table in its neutral state is Neon (Ne) ( atomic number (10) = no.of protons = no. of electrons) so neutral element with same electron configuration as O2- is Ne. Log in or register to post comments.

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According to the Pauli exclusion principle, when can two electrons occupy the same orbital?
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g in the following three compounds(1,2,3) arrange their relative reactivity towards the reagent CH3Cl / AlCl3. Justify your orde
Leto [7]

Answer:

3 > 2> 1

Explanation:

Aromatic compounds undergo electrophilic substitution reaction with several electrophiles.

Some substituted benzenes are more reactive towards electrophilic aromatic substitution than unsubstituted benzene.

Certain groups of substituents increase the ease with which an aromatic compound undergoes aromatic substitution.

If we look at the compounds closely, we will notice that only toluene leads to easy reaction with CH3Cl / AlCl3. Thus is due to the +I inductive effect of -CH3 which stabilizes the negatively charged intermediate produced in the reaction.

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3 years ago
Suppose the mole number of Ca2+ ions in a 50 mL water sample is quantified as 1.5 × 10−5 mol. What is the concentration of Ca2+
Anna11 [10]
\frac{1.5x 10^{-5} mol  Ca^{2+}}{50ml}  ( \frac{1 mol CaC O_{3} }{1 mol  Ca^{2+} } )( \frac{100,000 ppm}{1 molCaC O_{3}/L } ) \\ =0.0300234
5 0
3 years ago
Read 2 more answers
Calculate the cell potential at 25oC under the following nonstandard conditions: 2MnO4-(aq) + 3Cu(s) + 8H+(aq) ⟶ 2MnO2(s) + 3Cu2
baherus [9]

Answer:

1.346 v

Explanation:

1) Fist of all we need to calculate the standard cell potential, one should look up the reduction potentials for the species envolved:

(oxidation) Cu_{(s)} →Cu^{2+}_{(aq)} +2e E°=0.337 v

(reduction) MnO_{4 (aq)} + 3e + 4H^{+}_{(aq)}→MnO_{2 (aq)}+2H_{2}O E°=1.679 v

(overall) 2MnO_{4 (aq)}+3Cu_{(s)}+8H^{+}_{(aq)}→3Cu^{2+}_{(aq)}+2MnO_{2 (aq)}+4H_{2}O E°=1.342 v

2) Nernst Equation

Knowing the standard potential, one calculates the nonstandard potential using the Nernst Equation:

E=E^{0} -\frac{RT}{nF}Ln\frac{[red]}{[ox]}

Where 'R' is the molar gas constant, 'T' is the kelvin temperature, 'n' is the number of electrons involved in the reaction and 'F' is the faraday constant.

The problem gives the [red]=0.66M and [ox]=1.69M, just apply to the Nernst Equation to give

E=1.342 -\frac{298.15*8.314}{6*96500}Ln\frac{[.66]}{[1.69]}=1.346

E=1.346

5 0
3 years ago
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