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s2008m [1.1K]
1 year ago
13

Which of the following electron configurations represents an element that is MOST LIKELY to bond in a 1:1 ratio with calcium, Ca

?
1s 2 2s2 2p6 3s2 3p4

1s 2 2s2 2p6 3s2 3p5

1s 2 2s2 2p6 3s2 3p6

1s 2 2s2 2p6 3s2 3p2
Chemistry
1 answer:
fiasKO [112]1 year ago
7 0

There are two types of chemical compound one is covalent compound and another is ionic compound in chemistry. In ionic bonds, electrons are completely transferred. The correct option is option C.

<h3>What is chemical Compound?</h3>

Chemical Compound is a combination of molecule, Molecule forms by combination of element and element forms by combination of atoms in fixed proportion.

During formation of a compound calcium act as cation as Ca²⁺ so, if any element has to react with this cation in a 1:1 ratio, must have -2 charge over it . Option C represents S⁻², which react with Ca²⁺.

Therefore, the correct option is option C.

To learn more about chemical compound, here:

brainly.com/question/26487468

#SPJ1

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The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:
german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

We are given:

Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

  • <u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-0.337=0.462V

  • <u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.28)=1.079V

  • <u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.74)=1.539V

  • <u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.337-(-0.28)=0.617V

  • <u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=-0.28-(-0.74)=0.46V

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0
3 years ago
based on the article what are some of the advantages that can be given by radio frequency and microwaves​
bonufazy [111]

Answer:

higher data rates are

Explanation:

transmitted

as the bandwidth

is more

more antenna gain is possible

6 0
2 years ago
Write a balanced equation for the following neutralization reaction HBr+LiOH&lt;-&gt;
Sophie [7]

Answer:

HBr(aq) + LiOH(aq) = LiBr(aq) + H2O(l)

Explanation:

For this reaction, the reactants are the hydrobomic acid and the lithium hydroxide which produces the products lithium bromide and water.

8 0
3 years ago
A base can neutralize an acid. true or false
konstantin123 [22]

Answer:

false. acids do not neutralize bases.

Explanation:

4 0
3 years ago
Read 2 more answers
Write Lewis structures for the following molecules: (a) ICl, (b) PH3, (c) P4 (each P is bonded to three other P atoms), (d) H2S,
spin [16.1K]

Answer :  The Lewis-dot structure for the following molecules are shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

Now we have to determine the Lewis-dot structure for the following molecules.

(a) The given molecule is, ICl

As we know that iodine and chlorine have '7' valence electrons.

Therefore, the total number of valence electrons in ICl = 7 + 7 = 14

According to Lewis-dot structure, there are 2 number of bonding electrons and 12 number of non-bonding electrons.

(b) The given molecule is, PH_3

As we know that phosphorous has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in PH_3 = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

(c) The given molecule is, P_4

As we know that phosphorous has '5' valence electrons.

Therefore, the total number of valence electrons in P_4 = 4(5) = 20

According to Lewis-dot structure, there are 6 number of bonding electrons and 14 number of non-bonding electrons.

(d) The given molecule is, H_2S

As we know that sulfur has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in H_2S = 6 + 2(1) = 8

According to Lewis-dot structure, there are 4 number of bonding electrons and 4 number of non-bonding electrons.

(e) The given molecule is, N_2H_4

As we know that nitrogen has '5' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in N_2H_4 = 2(5) + 4(1) = 14

According to Lewis-dot structure, there are 10 number of bonding electrons and 4 number of non-bonding electrons.

(f) The given molecule is, HClO_3

As we know that chlorine has '7' valence electrons, oxygen has '6' valence electrons and hydrogen has '1' valence electrons.

Therefore, the total number of valence electrons in HClO_3 = 1 + 7 + 3(6) = 26

According to Lewis-dot structure, there are 12 number of bonding electrons and 14 number of non-bonding electrons.

(g) The given molecule is, COBr_2

As we know that bromine has '7' valence electrons, oxygen has '6' valence electrons and carbon has '4' valence electrons.

Therefore, the total number of valence electrons in COBr_2 = 4 + 6 + 2(7) = 24

According to Lewis-dot structure, there are 8 number of bonding electrons and 16 number of non-bonding electrons.

3 0
2 years ago
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