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DedPeter [7]
1 year ago
12

Hector claims that a single sodium ion and a single oxygen ion do not bond together to form a stable binary ionic compound. Is h

e correct? Why or why not?
A Hector is incorrect. The two ions have opposite charges, so they will bond together in a binary ionic compound.
B Hector is incorrect. The two ions have more than eighteen protons, so they will bond together in a binary ionic compound.
C Hector is correct. The sum of the charges of the two ions must be zero in order for them to form a stable ionic compound.
D Hector is correct. The sum of the protons of the two ions must be divisible by eight in order for them to form a stable ionic compound.
Chemistry
2 answers:
levacccp [35]1 year ago
8 0

Answer:

C

Explanation:

Looking at the periodic table, we can see that sodium is in group 1, so a sodium ion would be Na⁺, with a charge of +1. Oxygen is in group 16, so an oxygen ion would be O²⁻, with a charge of -2.

A compound formed only by a single sodium ion and a single oxygen ion would thus have a charge of -1, and in order to have a stable ionic compound its charge must be zero.

Anuta_ua [19.1K]1 year ago
5 0

Answer: A Hector is incorrect. The two ions have opposite charges, so they will bond together in a binary ionic compound.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Electronic configuration of sodium:

[Na]=1s^22s^22p^63s^1

Sodium atom will loose two electrons to gain noble gas configuration and form sodium cation with +1 charge.

[tex][Na^{+}]=1s^22s^22p^63s^0[/tex]

Electronic configuration of oxygen:

[O]=1s^22s^22p^4

Oxygen atom will gain one electron to gain noble gas configuration and form oxide ion with -2 charge.

[O^{2-}]=1s^22s^22p^6

Here sodium is having an oxidation state of +1 called as Na^+ cation and oxide O^{2-} is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Na_2O.

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

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lana66690 [7]

<u>Answer:</u> The element represented by M is Strontium.

<u>Explanation:</u>

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol

The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

8 0
1 year ago
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Magnesium burns in air with a dazzling brilliance to produce magnesium oxide: 2Mg (s) + O2 (g) →→ 2MgO (s) When 4.50 g of magnes
Klio2033 [76]

Answer:

7.46 g

Explanation:

From the balanced equation, 2 moles of Mg is required for 2 moles of MgO.

The mole ratio is 1:1

mole = mass/molar mass

mole of 4.50 g Mg = 4.50/24.3 = 0.185 mole

0.185 mole Mg will tiled 0.185 MgO

Hence, theoretical yield of MgO in g

mass = mole x molar mass

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How many structures are possible for a trigonal bipyramidal molecule with a formula of ax3y2?
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Check the attached file for the answer.

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Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

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Select all the true statements. Group of answer choices The +3 oxidation state is characteristic of the actinides. All actinides
Tanya [424]

Answer:

The +3 oxidation state is characteristic of the actinides.

All actinides are radioactive.

Cerium (Ce) rnakes 100th in abundance (by mass %).

The actinides are silvery and chemically reactive.

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