An aqueous solution of cobalt(II) fluoride, , is made by dissolving 6.04 grams of cobalt(II) fluoride in sufficient water in a 200. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of cobalt(II) fluoride in the solution

Answer:

37%

Explanation:

From the question, the equation goes does.

C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.

Mair=Mair/Rin

( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.

33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1

= 176/176+8

X= 0.37

0.37 × 100

X= 37%

Answer:

Mass PbCl₂ = 50.24g

Mass AgCl = 14.84g

Explanation:

The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:

Ag⁺ + Cl⁻ → AgCl(s)

Pb²⁺ + 2Cl⁻ → PbCl₂(s)

If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:

Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂

And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:

(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl

Moles of Cl⁻ that were added in the KCl solution are:

0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.

Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)

0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles

0.45409 + 0.00021X = 0.46464

0.00021X = 0.01055

X = 0.01055 / 0.00021

X = 50.24g

As X = Mass PbCl₂

And mass of AgCl = 65.08 - 50.24

The masses of the compounds in the precipitate can be found my knowing

the number of moles of chloride ion contributed by each compound.

Reasons:

The given parameter are;

Volume of KCl solution added = 0.242 L

Concentration of KCl solution = 1.92 M KCl

The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions

Precipitates formed = AgCl and PbCl₂

The mass of the precipitate = 65.08 g

Required:

The mass of PbCl₂ and AgCl in the precipitate

Solution;

Number of moles of chloride ions in a mole of PbCl₂ = 2 moles

Number of moles of chloride ions in a mole of AgCl = 1 mole

Let X represent the mass of PbCl₂ in the precipitate, we have;

The mass of AgCl in the precipitate = 65.08 g - X

[tex]\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}[/tex]

Number of moles of chloride ions from PbCl₂ is therefore;

[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}[/tex]

[tex]\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}[/tex]

[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-[/tex]

The number of moles of chloride ions from one mole of KCl = 1 mole

Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;

0.242 L × 1.92 moles/L = 0.46464 moles

Number of moles of chloride ions from KCl = 0.46464 moles

[tex]0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl[/tex]

Which gives;

[tex]\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}[/tex]

Therefore;

[tex]\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}[/tex]

The mass of PbCl₂ in the precipitate, X ≈ 49.24 g

The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g

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Answer:

Al(NO₃)₃: Acidic.

C₂H₅NH₃NO₃: Acidic.

NaClO: Basic

RbI: pH-neutral

CH₃NH₃CN: Solution basic

Explanation:

The general rules to determine if a solution is acidic, basic or neutral are:

For the salts:

Al(NO₃)₃. The repective acid is HNO₃ (Strong acid) and the base is Al(OH)₃ (Weak base). As the salt comes from strong acid and weak base. SOLUTION ACIDIC

C₂H₅NH₃NO₃. The acid is HNO₃ (Strong acid) and the base C₂H₅NH₃OH (Weak base). SOLUTION ACIDIC.

NaClO. Tha acid is HClO (weak acid), and the base NaOH (Strong base). SOLUTION BASIC.

RbI: The acid is HI (Strong acid) and the base RbOH (Strong base). pH-NEUTRAL

CH₃NH₃CN. The acid is HCN (weak acid; pKb = 4.79) and  the base CH₃NH₃OH (weak base; pKa = 10.64). Both weak acid and base will produce each hydrolisis. The lower pK will predominate. That is the weak acid. SOLUTION BASIC

Solution of Al(NO₃)₃ and C₂H₅NH₃NO₃ salts is acidic, NaClO is basic and of RbI & CH₃NH₃cyanide is neutral in nature.

pH of any solution tells about the acidity or basicity of the solution, pH of any solution ranges from 0 to 14 and from acidity to basicity.

Hence, appropriate differentiation was done above.

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Answer:

Explanation:

a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .  

C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O

O is provided by KMnO₄

b ) In this reaction isophthalic acid is formed .

C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂

c)

4-Propyl-3-t-butyltoluene

In this oxidation , three side chains of ring  are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .

The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )

Answer:

Option B. S

Explanation:

All of the options except sulphur, S is metal.

Metals tend to lose electron in order to form ion. Non metals on the other hand gain electron to form ion.

Sulphur, S has atomic number of 16 with electronic configuration as:

S (16) => 1s² 2s²2p⁶ 3s²3p⁴

From the above illustration, we can see that sulphur needs two more electrons to complete it's octet configuration.

Therefore, sulphur, S will gain two electrons to form ion.

As stated earlier, the rest option given are all metals which will form ion by losing electron(s).

Answer

B) Sulphur (S)

Explanation

Here in the options we have been provided with elements like Nickel (Ni), Sulphur (S), Sodium (Na), Chromium (Cr) and Beryllium (Be) but except for Sulphur all the other ones are metals.

Now, let us understand what is a metal and a non-metal.

So, sulphur being the only non metal will accept electron to complete its octate and to stablize itself and form a Anion.

Now let us also look at the electronic configuration of Sulphur to get the picture more clearly

        [tex]S\rightarrow 1s^2\; 2s^2\;2p^6\;3s^2\;3p^4[/tex]

so here the p-subshell is incomplete and is in need of 2 electrons.

Therefore the element which is most likely to gain electrons, forming an Anion will be sulphur.

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Answer:

Isomers are molecules that have the same molecular method, however have a unique association of the atoms in space. That excludes any extraordinary preparations which can be sincerely because of the molecule rotating as an entire, or rotating about precise bonds.

Answer:

Sb

Explanation:

The periodic trend for atomic radius is that it decreases from left to right and increases from top to bottom, therefore the elements with the larger atomic radius will be the ones which are closest to the bottom left corner of the periodic  table. Since all of these elements are in the same group, the one with the largest atomic radius will be the one at the "bottom", and that is Sb.

Answer:

Coordinate or r = (7,9).

Explanation:

Data obtained from the question include the following:

Mid point = (8,7)

Coordinate of S = (9,5)

Coordinate of r =...?

We shall determine the coordinate of r as follow:

Let the coordinate of r be (x2, y2)

Mid point = (x1 + x2)/2 , (y1 + y2)/2

Mid point = (8,7)

Coordinate of S = (9,5)

x1 = 9

y1 = 5

x2 =?

y2 =?

The value of x2 can be obtained as follow:

8 = (x1 + x2)/2

8 = (9 + x2)/2

Cross multiply

9 + x2 = 2 × 8

9 + x2 = 16

Collect like terms

x2 = 16 – 9

x2 = 7

The value of y2 can be obtained as follow:

5 = (y1 + y2)/2

7 = (5 + y2)/2

Cross multiply

5 + y2 = 2 × 7

5 + y2 = 14

Collect like terms

y2 = 14 – 5

y2 = 9

Coordinate of r = (x2, y2)

Coordinate or r = (7,9)

Answer:

1. REDOX

2. None of the above

3. Precipitation

4. Preicipitation

5. Acid base neutralization

Explanation:

Reactions where a solid is formed, are named as precipitation. This solid is called precipitated.

Option 4 and 3.

3) CaO (s) + CO₂ (g) →  CaCO₃(s)

4) KOH (aq)  + AgCl (aq)  →  KCl (aq)  + AgOH(s)

Reactions where water is produced, and you have an acid and a base as reactants, are named as neutralization. You called them acid-base because, the products.

5) Ba(OH)₂ (aq)  +  2HNO₂(aq)  →  Ba(NO₂)₂ (aq) + 2H₂O(l)

Redox, are the reactions where one of the reactans can be oxidized and reduced, when a mole of electrons is released, or gained.

1) Ba(ClO₃)₂ (s)  → BaCl₂ (s) +  3O₂(g)

Oxygen from the chlorate is oxidized (increases the oxidation state from -2 to 0) and the chlorine is reduced (decreases the oxidation state from +5 to -1).

2.  2NaCl(aq)  +  K₂S(aq)  Na₂S (aq)  + 2KCl (aq)

None of the above

Answer:

Aromatic Hydrocarbons

Explanation:

Aromatic (Pleasant Odour) Hydrocarbons are those having pleasant odours.

Answer:

substituted hydrocarbons

Explanation:

i think

Answer:

too low

Explanation:

If our aim is to recover the naphthalene and measure its mass after separation, then we must not allow any vapour to escape.

Naphthalene is a sublime substance, it can be separated by sublimation. It changes directly from solid to gas. This vapour must be kept securely so that none of it escapes. If part of the naphthalene vapour happens to escape accidentally, then the measured mass of naphthalene will be too low compared to the mass of naphthalene originally present in the mixture.

Answer:

Explanation:

a) 2 chloro butane

b) 2-3 dimethyl butane

c) 2 bromo 3 nitro pentane

d) 2-3 trimethyl pentane

e) 2-bromo,3-methyl,4-nitro hexane

f) 2-methyl cyclo butane

Answer:

1-cyclopentylhexan-2-one

Explanation:

1-cyclopentylhexan-2-one

Answer:

Do not conduct electricity as solids.

Explanation:

Hello,

In this case, we should remember that salts are formed when an acid and base react in order to yield the salt and water due to the ions exchange during neutralization chemical reactions. For instance, when hydrochloric acid  (acid) reacts with potassium hydroxide (base), sodium chloride (salt) and water are yielded via:

[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]

Moreover, it is widely known that salts are formed by electrovalent/ionic bonds which involves electron transfer so the metallic atom becomes positively charged (cation) whereas the non-metallic atom becomes negatively charged (anion) once the electrons are received so it can conduct electricity when dissolved in water yet not when solid since electron transfer is facilitated by the aqueous media, otherwise, ions remain together. Thereby, answer is do not conduct electricity as solids.

Regards.

Answer:

c

Explanation:

Answer:

See explanation

Explanation:

A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.

Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.

The canonical structure of the carbonate ion as well as the lewis structure of phosphine is shown in the image attached to this answer.

Answer:

A

Explanation:

The number of protons indicates the element so we know it's potassium. To get the number of neutrons you subtract the number of protons (19) from the mass number which for potassium is 39.

39-19=20 neutrons

Because you have 18 neutrons then yours would be an isotope.

Answer: A. Isotope of potassium(K)

Explanation: Founders Educere answer.

Answer:

0.1 M

Explanation:

The overall balanced reaction equation for the process is;

IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)

Generally, we must note that;

1 mol of IO3^- require 6 moles of S2O3^2-

Thus;

n (iodate) = n(thiosulfate)/6

C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6

Concentration of iodate C(iodate)= 0.0100 M

Volume of iodate= V(iodate)= 26.34 ml

Concentration of thiosulphate= C(thiosulfate)= the unknown

Volume of thiosulphate=V(thiosulfate)= 15.51 ml

Hence;

C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)

0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M

Answer:

The correct option is d

Explanation:

Nuclide is synonymous with groups of electrons or protons, that is, a nuclide is the grouping of nucleons.

Answer:

b) rate = kAB.

Explanation:

Hello,

In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:

[tex]A+B \rightarrow AB\ \ (slow)[/tex]

In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:

[tex]rate=k[A][B][/tex]

Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.

Best regards.

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T}[/tex]

We can solve for the temperature as follows:

[tex]T=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]

Thus, with the proper units, we obtain:

[tex]T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C[/tex]

Hence, answer is approximately 100 °C.

Best regards.

The given question is incomplete. The complete question is given here :

The equilibrium constant for the reaction is [tex]1.1\times 10^6[/tex] M.

[tex]HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)[/tex]

This value indicates that

A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

B. HCN is a stronger acid than HONO

C. The conjugate base of HONO is [tex]ONO^-[/tex]

D. The conjugate acid of CN- is HCN

Answer: A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

Explanation:

Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.

When [tex]K_{p}>1[/tex]; the reaction is product favoured.

When [tex]K_{p};<1[/tex] ; the reaction is reactant favored.

[tex]When K_{p}=1[/tex]; the reaction is in equilibrium.

As, [tex]K_p>>1[/tex], the reaction will be product favoured and as it is a acid base reaction where [tex]HONO[/tex] acts as acid by donating [tex]H^+[/tex] ions and [tex]CN^-[/tex] acts as base by accepting [tex]H^+[/tex]

Thus [tex]HONO[/tex] is a strong acid thus [tex]ONO^-[/tex] will be a weak conjugate base and [tex]CN^-[/tex] is a strong base which has weak [tex]HCN[/tex] conjugate acid.

Thus the high value of K indicates that [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]

Answer:

The final temperature of sulfur dioxide gas is 215.43 C

Explanation:

Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.

Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:

[tex]\frac{P}{T}=k[/tex]

Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:

[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]

The reference temperature is the absolute temperature (in degrees Kelvin)

In this case:

Replacing:

[tex]\frac{0.450atm}{293.15 K} =\frac{0.750 atm}{T2}[/tex]

Solving:

[tex]T2 =\frac{0.750 atm}{\frac{0.450atm}{293.15 K} }[/tex]

[tex]T2=\frac{0.750 atm}{0.450 atm} *293.15K[/tex]

T2=488.58 K

Being 273.15 K= 0 C, then 488.58 K= 215.43 C

The final temperature of sulfur dioxide gas is 215.43 C

Answer:

(a) [tex]Ksp=4.50x10^{-7}[/tex]

(b) [tex]Ksp=1.55x10^{-6}[/tex]

(c) [tex]Ksp=2.27x10^{-12}[/tex]

(d) [tex]Ksp=1.05x10^{-22}[/tex]

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) [tex]BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)[/tex]

[tex]Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}[/tex]

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

[tex]Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}[/tex]

(B) [tex]Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)[/tex]

[tex]Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}[/tex]

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

[tex]Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}[/tex]

(C) [tex]NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)[/tex]

[tex]Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}[/tex]

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

[tex]Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}[/tex]

(D) [tex]La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)[/tex]

[tex]Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}[/tex]

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

[tex]Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}[/tex]

Best regards.

Answer:

Explanation:

The substances are:

-) Ethane, [tex]C_2H_6[/tex]

-) 1-pentanol, [tex]C_5H_1_1OH[/tex]

-) Potassium chloride, [tex]KCl[/tex]

-) Propane,  [tex]C_3H_8[/tex]

For this question, we have to remember the structure of water. Due to the electronegativity difference between oxygen and hydrogen in this structure, we will have the formation of dipoles. The dipoles interact better with net charges, due to this, the Potassium chloride is the compound with highest solubility (due to the formation of a cation and an anion):

[tex]KC~l->~K^~+~Cl^-[/tex]

Then, in 1-pentanol we an "OH". This structure due to the presence of the hydroxyl group can form hydrogen bonds. Therefore,  this compound would be the second more soluble.

Finally,  the difference between propane and ethane is a carbon. In propane, we have an additional carbon. If we have more carbons we will have more area of ​​interaction. If we have more area we will have more solubility therefore propane is more soluble than ethanol.

In conclusion, the rank from most soluble to least soluble is:

1) Potassium chloride, [tex]KCl[/tex]

2) 1-pentanol, [tex]C_5H_1_1OH[/tex]

3) Propane,  [tex]C_3H_8[/tex]

4) Ethane, [tex]C_2H_6[/tex]

I hope it helps!

Order of solubility in water will be:

KCl > C₅H₁₁OH > C₃H₈ > C₂H₆

Any solvent soluble in water due to its polarity and ability to form hydrogen bonds. The presence of hydrogen bonding between molecules of a substance indicates that the molecules are polar. This means the molecules will be soluble in a polar solvent such as water.

Substances that are given:

Ethane(C₂H₆), 1-pentanol(C₅H₁₁OH), Potassium chloride(KCl) and propane(C₃H₈).

We will look at each compound one by one:

Order of solubility in water will be:

KCl > C₅H₁₁OH > C₃H₈ > C₂H₆

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Answer:

The Nuclear magnetic resonance is the process this technique does not use radiation.

The  ms is an sensitive technology can be a massive number and small sample of the blood.

Explanation:

The Nuclear magnetic resonance we look at the both side of that coin.

The technique provides that fatty acid composition and various including amino acids.

These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-

(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.

The MS is an extremely sensitive technology using a very small number of the blood.

(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume

MS can be fine mapping metabolic pathways to sign analytical strategy.

Answer:

F

Explanation:

Halogens may interact with the benzene ring via inductive or resonance effects. Halogens deactivate the benzene ring by inductive effect rather than by resonance effects.

The lone pairs of electrons present on the halogen atoms may be donated to the ring by resonance, but an opposite effect, the inductive pull (-I inductive effect) of the halogen atoms on electrons away from the benzene ring due to the high electro negativity of the halogens leads to a deactivation of the ring towards electrophilic substitution.

Hence inductive electron withdrawal by the halogen atom predominates over electron donation by resonance effect and the benzene ring g is deactivated towards electrophilic substitution at the ortho and para positions.

Radiologists are medical doctors that treat injuries using medical imaging (radiology)

Answer:

a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.

Explanation:

Answer:

For carbon, the effective nuclear charge is 3.25 and the valence electrons will reside in the orbitals 2s^2 and 2p^2

For silicon, the effective nuclear charge is 4.15 and its valance electrons will reside in the orbitals 3s^2 and 3p^2

Explanation:

Carbon

The effective nuclear charge of carbon is 3.25

To get the orbitals in which it’s valence electron reside, let’s write the electronic configuration

The atomic number of carbon is 6

So the configuration will be;

1s^2 2s^2 2p^2

So the valence electrons will reside in the orbitals 2s^2 and 2p^2

For silicon;

It’s effective nuclear charge is +4.15

The electronic configuration of silicon with atomic number 14 is;

1s^2 2s^2 2p^6 3s^2 3p^2

So the valence electrons will reside in the orbitals 3s^2 and 3p^2