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3 years ago

What is the chemical reaction for Mg + NaOH also with the ionic equations and subscripts?

1 answer:

5 0

The chemical reaction of magnesium and sodium hydroxide would yield magnesium hydroxide and sodium. The chemical reaction is expressed as:
Mg(s)+ 2NaOH(aq)→Mg(OH)2(s)+2Na
In ionic form,
Mg(s) + 2Na+ + 2OH−(aq)→Mg(OH)2(s) + 2Na

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Some patterns of electron configuration are listed below. In each case, [G] stands for a noble-gas core and n, m, or o stand for

Nonamiya [84]

Answer:

G]ns^2np^5 group 17 (p-block)

G]ns^2np^2 group 14 (p-block)

G]ns^2mf^14 group 16 (f-block)

Explanation:

The outermost electronic configuration of an element shows the group to which it belongs in the periodic table as shown above in the answer. In addition, to that, we can be able to know from its electronic configuration, whether the element is a metal or not.

For instance;

G]ns^2mf^14 is a rare earth metal, G]ns^2np^2 group 14 is a metalloid while G]ns^2np^5 group 17 is a nonmetal.

4 0

3 years ago

Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of

Alenkinab [10]

Answer:

Dark matter makes up 85% of the mass of the universe. Dark matter is not directly observable because it doesn't interact with any electromagnetic wave. In the development of the universe, without dark matter, the universe will not function, move or rotate as it does now (this speculation led to the quest to find the anomaly of mass and energy in the known universe, eventually leading to the idealization of dark matter) and will not have enough gravitational force to hold it together. After the big bang, the presence of dark matter and energy ensured that the newly formed universe didn't just float away, rather, it provided enough gravitational force to hold the universe while still allowing it to expand sufficiently.

The development of the universe would have been different without the universe in the sense that the young universe won't have enough mass to hold it together, and the universe would have simply floated apart. The behavior of the universe would have been different from what we observe now, and some physical laws that applies now will not apply to the universe.

6 0

3 years ago

Estimate the coordination number for the cations in each of these ceramic oxides and also the coordination numbers of the oxygen

Juli2301 [7.4K]

Answer:

CN cation, anion ] respectively are thus [6,4] , [2,3] ,[6,6] , [6,6] , [6,3] , [12,6,2] , [6], [6].

Explanation:

The coordination number CN is the number of ligand atoms bonded (coordinate bonds) directly to the central of the metal ion. It is not the same as the oxidation state of the metal ion or complex.

Coordination number – the number of anions surrounding the cation.

In solving for CN we need to understand Pauling's rules.

According to Linus Pauling, 1932

“Pauling’s rules” for crystal structures, makes assumptions for ionic bonding. It states that ionic structure is understood using electrostatic rules of attraction and repulsion.

Cations and anions surround each other to neutralize charge – and these one can rationalize crystal structure with coordination number.

Ratio of cationic/anionic radius

• The structure of D-Al2O3 results in coordination number of 6 and 4 for cation and anion respectively.

• The average oxygen coordination number in v-B2O3 is equal to the average cation coordination number × cation/anion ratio (2/3).

• Co-ordination number of Ca2+ ion is =6;

In CaO crystal, Ca2+ is a cation and O2- is an anion. Cationic (Ca2+) has radius 100 pm and anionic (O2-) has radius 145 pm.

Ratio of cationic/anionic radius is:

r⁺/r⁻ = 100 / 145

r⁺/r⁻ = 0.69

CaO will form FCC lattice.

Coordination number in FCC lattice is 6. Therefore CN of Ca2+ = 6.

For MgO:

r Mg2+/ r O2- = 86pm / 126 pm =0.683

The cordination number for the cation is 6. MgO with ions Mg+2 and O-2 will have a AX type stochiometry exhibiting the

crystal structure of sodium chloride.

For TiO2:

The CN of the titanium (IV) cation is 6, which is twice the CN of the oxide anion, which is 3.

This fits with the formula unit of TiO2, since there are twice as many O2− ions as Ti4+ ions.

Consequently the crystal structure of all ionic compounds reflects the formula unit.

For LaAlO3 a Cubic perovskites (ABX3)

In perovskite structures, B cations are coordinated by six X anions, while A

cations present CN = 12 (also coordinated by X anions). The X anions have CN = 2, being

coordinated by two A cations, since the distance A-O is about 40% larger than the B-O

bond distance. The correct ionic radii (rA, rB, rX), taken from one of Shannon’s work.

rA = 1.36 pm

rB = 0.535 pm

rX = 1.35 pm

Forsterite Mg2SiO4

We have mixed sites of Si4+ in tetrahedral site, Mg2+ in octahedral site, O atoms anions .

They all forms octahedral chains/strips. The CN is estimated to (6) octahedral, with an average ratio 0.414

Nickel Cobaltite Ni(Co2O4)3

bidentate ligand includes the Oxalate, three oxalate ligands form six-coordinate bonds around the Ni2+ ion.

Co-ordinate number of Nickel in [Ni(C2O4)3] 4− is 3×2=6.

5 0

3 years ago

The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed

Oksanka [162]

Answer:

(a) The rate of disappearance of $O_{2}$ is: $4.65*10^{-5}$ M/s

(b) The value of rate constant is: 0.83036 $M^{-2}s^{-1}$

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

$2NO(g)+O_{2}->2NO_{2}$

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = $-\frac{d}{dt}[O_{2}]$ = $\frac{1}{2}\frac{d}{dt}[NO_{2}]$ -----(1)

According to the question, the reaction is second order in NO and first order in $O_{2}$ .

Then we can say that, rate = k $[NO]^{2}[O_{2}]$ -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

$-\frac{d}{dt}[NO]$ = $9.3*10^{-5}$ M/s.

(a) From (1), we can get the rate of disappearance of $O_{2}$ .

Rate of disappearance of $O_{2}$ = $-\frac{d}{dt}[O_{2}]$ = (0.5)*( $9.3*10^{-5}$ ) M/s = $4.65*10^{-5}$ M/s.

(b) The rate of the reaction can be obtained from (1).

rate = $-\frac{1}{2} \frac{d}{dt}[NO]$ = (0.5)*( $9.3*10^{-5}$ )

rate = $4.65*10^{-5}$ M/s

The value of rate constant can be obtained by using (2).

rate constant = k = $\frac{rate}{[NO]^{2}[O_{2}]}$

k = $\frac{4.65*10^{-5}}{(0.040)^{2}(0.035)}$ = 0.83036 $M^{-2}s^{-1}$

k = $\frac{rate}{[NO]^{2}[O_{2}]}$

Substituting the units of rate as M/s and concentrations as M, we get:

$\frac{Ms^{-1} }{M^{3}}$ = $M^{-2}s^{-1}$

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

$rate\alpha [NO]^{2}$

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of $(1.8)^{2}$ = 3.24

5 0

3 years ago

What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?

My name is Ann [436]

Answer:

602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The yield is measured by the amount of each product produced by the reaction.

The chemical formula of fluorene is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The oxidation, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

$2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O$

To calculate the yield follow these steps:

1. Mole ratio

$2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O$

2. Convert 175mg of fluorene to number of moles

$175mg/times 1g/1,000mg=0.175g$

Number of moles = mass in grams / molar mass

$\text{number of moles}=0.175g/166.223g/mol=0.0010528mol$

3. Set a proportion for each product of the reaction

a) For CO₂

i) number of moles

$2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x$

$x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2$

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

mass = number of moles × molar mass

mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) For H₂O

i) number of moles

$0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O$

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

mass = number of moles × molar mass

mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg

4 0

3 years ago