11. In TLC analysis of ferrocene and acetylferrocene (on silica TLC plate) which prediction is correct: A) ferrocene is more polar and moves higher up the plate (higher Rf value) B) Acetylferrocene is more polar and moves higher up the plate (higher Rf value) C) ferrocene is less polar and moves higher up the plate (higher Rf value) D) Acetylferrocene is less polar and moves higher up the plate (higher Rf value)

Answer:

This description shows a methyl group.

Explanation:

Answer:

3 fluorine atoms will be present

Answer:

3

Explanation:

The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.

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Answer:

The given statement is false.

Explanation:

So that the given is incorrect.

Answer: B. The period number tells which is the highest energy level occupied by the electrons

Explanation:

Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.

hence, the correct option is B. The period number tells which is the highest energy level occupied  by the electrons.

The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.

The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.

When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.

Therefore, option B , second option is correct.

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1. 21.67g/ml

2. aluminium

Explanation:

1. density = mass/volume

385.8/17.8= 21.67ml

2. 1g/ml=0.1g/cm^3

21.67g/ml = 2.167g/cm^3

..... substance is probably aluminium

1.  the density of the metal is 21.67g/ml

2.  This metal is most likely aluminum

1.

[tex]density = mass \div volume[/tex]

[tex]385.8\div 17.8= 21.67ml[/tex]

2.

1g/ml=0.1g/cm^3

So,  

21.67g/ml = 2.167g/cm^3

Therefore,  substance is probably aluminum

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Answer:

we wear cotton clothes because it helps to cool us down and remove the excess heat that causes us to feel hot.

Answer:

[tex]\boxed{\mathrm{view \: explanation}}[/tex]

Explanation:

We wear cotton clothes in the summer beacuse cotton absorbs and removes body moisture caused by the sweat and allows better air circulation than fabric clothes.

Answer:

1. 0.138g of valium would be lethel in the woman

2. 125mg/min is the drip of the patient

Explanation:

1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.

A person that weighs 200lb requires:

200lb × (453.6g / 1lb) × (1kg / 1000g) = 90.72kg (Weight of the woman in kg)

90.72kg × (1.52mg / kg) =

137.9mg ≡

2. The IV contains 1.5g = 1500mg/mL.

If the patient is receiving 5.0mL/h, its rate in mg/h is:

5.0mL/h × (1500mg/mL) = 7500mg/h

Now as 1h = 60min:

7500mg/h × (1h / 60min) =

Answer:

1. Increased levels of fructose-2,6-bisphosphatase : Activate gluconeogenesis Inhibit glycolysis

2. Activation of fructose-2,6-bisphosphate (FBPase-2) : Activate glycolysis Inhibit gluconeogenesis

3. Increased glucagon levels : Activate gluconeogenesis Inhibit glycolysis

4. Activation of PFK-2 : Activate glycolysis Inhibit gluconeogenesis

5. Increased levels of CAMP : Activate gluconeogenesis Inhibit glycolysis

Explanation:

Glycolysis is the breakdown of glucose molecules in order to release energy in the form of ATP in response to the energy needs of the cells of an organism.

Gluconeogenesis is the process by which cells make glucose from other molecules for other metabolic needs of the cell other than energy production.

Glycolysis and gluconeogenesis are metabolically regulated in the cell by various enzymes and molecules.

The following shows the various regulatory methods and their effects on both processes:

1. The enzyme fructose-2,6-bisphosphatase functions in the regulation of both processes. It catalyzes the breakdown of the molecule fructose-2,6-bisphosphate which is an allosteric effector of two enzymes phosphofructokinasse-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase-1 which fuction in glycolysis and gluconeogenesis respectively.

Increased levels of fructose-2,6-bisphosphatase  activates gluconeogenesis and inhibits glycolysis by its breakdown of fructose-2,6-bisphosphate.

2. Fructose-2,6-bisphosphate increases the activity of PFK-1 and inhibits the the activity of FBPase-1. The effect is that glycolysis is activated while gluconeogenesis is inhibited.

3. Glucagon is a hormone that stimulates the synthesis of cAMP. It fuctions to activate gluconeogenesis and inhibit glycolysis.

4. Phosphosfructikinase-2, PFK-2 is an enzyme that catalyzes the formation of fructose-2,6-bisphosphate. Activation of PFK-2 results the activation of glycolysis and inhibition of gluconeogenesis.

5. Cyclic-AMP (cAMP) synthesis in response to glucagon release serves to activate a cAMP-dependent protein kinase which phosphorylates the bifunctional protein PFK-2/FBPase-2. This phosphorylation enhances the activity of FBPase-2 while inhibiting the activity of PFK-2, resulting in the  activation of gluconeogenesis and inhibition of glycolysis.

Answer:B

Explanation:

Answer: i think it is c

Explanation: i checked my textbook.

Answer:

-179.06 kJ

Explanation:

Let's consider the following balanced reaction.

HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)

We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))

ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)

ΔH°r = -179.06 kJ

Answer:

c iodine

Explanation:

fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride

The Average atomic mass of phosphorus is 29.9.

The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.

Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.

An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.

The versions of an element with different neutrons have different masses and are called isotopes.

The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,

Average atomic mass of P = ∑(Isotope mass) (its abundance)

∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)

Abundance of isotope = % of the isotope / 100.

∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.

Hence , The Average atomic mass of phosphorus is 29.9.

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Answer:

The correct answer is 51.2 degree C.

Explanation:

The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.  

Now the dHreaction = dHf (product) - dHf (reactant)  

= -909.3 - (-814)

dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.  

The heat change in calorimetry can be determined by using the formula,  

q = mass * specific heat capacity * change in temperature -----------(i)

Based on the given information, the density of H₂SO₄ is 1.060 g/ml

The volume of H₂SO₄ is 1 Liter

Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams

The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.  

ΔT = T₂ -T₁ = T₂ - 298.2 K

Now putting the values in equation (i) we get,  

95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)

(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K

T₂ = 298.2 K + 26 K

T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.  

Answer:

5 .04 percent.

Explanation:

it is also known as sulfur dioxide. so it's 5.0.4 percent.

Your question is incomplete. Read below to find the content.

0.4 % is the percent composition of sulfur in SO2.

In the laboratory, sulfur dioxide is prepared by the reaction of metallic sulfite or a metallic bisulfite with dilute acid. For example, a reaction between the dilute sulphuric acid and sodium sulfite will result in the formation of SO2.

The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100. Here, the quantity is measured in terms of grams of the elements present.

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Answer:

The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.

Explanation:

1s^2

2s^2

2p^6

3s^2

3p^6

4s^2

3d^10

4p^5

Answer:

The reaction given in the question is:  

2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)

The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:

ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.  

The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.  

Now the values of ΔH° and ΔS° are,  

ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)

ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)

ΔH°rxn = -1277.56 kJ/mole

ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)

ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15

ΔS°rxn = 313.11 J/mole/K

Now the formula for calculating ΔG°rxn is,  

ΔG°rxn = ΔH°rxn - TΔS°rxn

ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole

ΔG°rxn = -1370.86 kJ/mol

Answer:

2NaOH(s) → Na₂O(s) + H₂O(g)

Hope that helps.

Answer:

ESCALAS MAYORES (D, E, G, A, B) Porfavor necesito ayuda,te lo agradecería muchísimo!!

Es urgente

Answer:

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

Explanation:

Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),

Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of  48  electrons surrounding its nucleus.

The electronic configuration of Cadmium is;

Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10

The shorthand notation is given as;

Cd: [Kr] 4d¹⁰5s²

Cd2+ means that it has two less electrons, hence it's electron configuration is given as;

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

Answer:

0.641 g of Nitrogen are present in the mixture.

Explanation:

We use the Ideal Gases Law, to solve this question.

For the mixture:

P mixture . V mixture = mol mixture . R . T

We convert the T° to K →  23°C + 273 = 296 K

R = Ideal gases constant → 0.082 L.atm/mol.K

1 atm . 2L = mol mixture . 0.082 L.atm/mol.K  . 296K

2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles

We know that sum of partial pressure = 1

Partial pressure N₂ + Partial pressure O₂ = 1

1 - 0.722 atm = Partial pressure N₂ → 0.278 atm

We apply the mole fraction concept:

Partial pressure N₂ / Total pressure = Moles N₂ / Total moles

Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles

Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles

We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g

641 mg

Answer:

The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3

Explanation:

To get the molar concentration of a solution we will use the formula:

Molar concentration = mass of HCl/ molar mass of HCl

Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.

We can extract the mass of the solution from its density which is 1.2g/mL

We will further perform our analysis by considering only 1 ml of this aqueous solution.

The mass of the substance present in this solution is 1.2g.

The mass of HCl Present is 40% of 1.2 = 0.48 g.

The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol

Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3

Answer:

The correct answer is option B and D, that is, halogen (chlorine) and hydroxyl.

Explanation:

An artificial sweetener and sugar substitute is sucralose. It is noncaloric as the majority of the sucralose ingested does not get dissociated within the body. The generation of sucralose takes place by the chlorination of sucrose. It is about 300 to 1000 times sweeter in comparison to sucrose.  

The consumption of sucralose is safe for both nondiabetics and diabetics, it is used in various food and beverage components due to non-caloric sweetener characteristics. It does not affect the levels of insulin and does not affect dental health. As it is produced by chlorination of sucrose, thus, the functional groups present in it are a halogen (chlorine) and a hydroxyl.  

Answer:

CH3CH2C≡CH

Explanation:

The particular reaction under study is known as the alkykation of acetylide ions. An acetylide ion can be alkykated using a suitable alkyl halide. The overall scheme of the reaction is;

CH≡C^- + RX -----> RC≡CH + X^-

This reaction is most effective when primary alkyl halides are used. It involves SN2 substitution of a halide in the alkyl halide by an acetylide ion. Secondary, tertiary or even bulky primary substrates are known to yield alkenes and alkynes owing to elimination by E2 mechanism.

Answer:

1. EF = PSCl₃; 2. MF = PSCl₃  

Explanation:

1. Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our first job is to calculate the molar ratio of P:S:Cl.

Assume 100 g of the compound.

(a) Calculate the mass of each element.

Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.

(b) Calculate the moles of each element

[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]

(c) Calculate the molar ratio of the elements

Divide each number by the smallest number of moles

P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3

(d) Write the empirical formula

EF = PSCl₃

The empirical formula for this compound is PSCl₃.

2. Molecular formula

(a) Calculate the ratio of the molecular and empirical formula masses

n = (169.4 u)/(169.40 u) = 1.000 ≈ 1

(b) Calculate the molecular formula

MF = (EF)ₙ = (EF)₁ = PSCl₃

The molecular formula for this compound is PSCl₃.

Answer:

[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]

ΔG ≅ 199.91 kJ

Explanation:

Consider the reaction:

[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]

[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]

[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]

[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]

[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]

[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]

The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :

[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]

[tex]K_p = \dfrac{1}{ (22.20)}[/tex]

[tex]K_p[/tex] = 0.045

[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]

where;

R = gas constant = 8.314 × 10⁻³ kJ

[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]

[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]

[tex]\Delta G =207.6+ (-7.687048037)[/tex]

[tex]\Delta G =[/tex] 199.912952  kJ

ΔG ≅ 199.91 kJ

Answer:

28.0mL of the 0.0500M NaOH solution

Explanation:

0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.

The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:

H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O

Where 1 mole of the acid reacts per mole of the base.

You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.

the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:

0.126g ₓ (1mol / 90.08g) = 1.40x10⁻³ moles of H₃C-CH(OH)-COOH

To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:

1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =

Answer:

1. oxygen; carbonyl; nitrogen; α-helices or β-pleated sheets

2. carbonyl; amino; the next turn of the helical chain; carbonyl; amino; parallel sections of a long polypeptide chain.

Explanation:

The secondary structure of proteins is of two major conformations, the α helix and β conformations which are very stable.

1. The oxygen atoms of the carbonyl groups and the hydrogen atoms attached to the nitrogen atoms form α-helices or β-pleated sheets.

2. In the α-helix, hydrogen bonds form between the carbonyl oxygen atom and the amino hydrogen atom of a peptide bond in the next turn of the helical chain. In the β-pleated sheet, hydrogen bonds form between the carbonyl oxygen atom and the amino hydrogen atom of a peptide bond in parallel sections of a long polypeptide chain.

Answer:

The correct answer is statement A.

Explanation:

In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.  

For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.  

The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.

In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.

The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.

TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.

Thus, option A is correct.

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Answer:

a. 1

b. 0.465M NaOH

Explanation:

KHP reacts with NaOH as follows:

KHP + NaOH → KP⁻ + Na⁺ + H₂O

a. Molar ratio represents how many moles of NaOH reacts per mole of KHP. As you can see in the reaction, 1 mole of NaOH reacts with 1 mole of KHP. Molar ratio is:

1/1 = 1

b. With volume and molar concentration of the KHP solution you can find how many moles of KHP were added until equivalence point, thus:

15.5mL = 0.0155L ₓ (0.600 moles KHP / L) = 0.0093 moles of KHP

In equivalence point, moles of NaOH = Moles KHP. That means moles of NaOH titrated are 0.0093 moles NaOH.

The volume of the NaOH solution was 20mL = 0.020L. Molarity of the solution is:

0.0093 moles NaOH / 0.020L =

a. The balanced equation shows a 1:1 molar ratio between NaOH and KHC₈H₄O₄. This means that for every 1 mole of NaOH, we require 1 mole of KHC₈H₄O₄. Therefore, the molar ratio of NaOH:KHC₈H₄O₄ is 1:1.

The balanced equation for the reaction:

NaOH + KHC₈H₄O₄ → NaKC₈H₄O₄ + H₂O

b. Molarity of KHP solution × volume of KHP solution = Molarity of NaOH solution × volume of NaOH solution at the equivalence point

Molarity of KHP solution = 0.600 M

Volume of KHP solution = 15.5 mL = 0.0155 L

Volume of NaOH solution at the equivalence point = 20 mL = 0.0200 L

Molarity of NaOH solution = (Molarity of KHP solution × volume of KHP solution) / volume of NaOH solution at the equivalence point

Molarity of NaOH solution = (0.600 M × 0.0155 L) / 0.0200 L

Molarity of NaOH solution ≈ 0.465 M

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Answer:

Explanation:

   In a triangle

a / sin A = b / sinB = c / sinC

Putting the values

43 / sin A = 44 / sinB

sinA / sinB = 43 / 44 = 1 / 1.023

A + B = 180 - 83 = 97

sinA / sin ( 97 - A ) = 1 / 1.023

sin 97 cos A - cos 97 sin A = 1.023 sin A

= .9925 cos A + .122 sin A = 1.023 sin A

.9925 cos A = .901 sin A

squaring

.985 cos²A = .8118 sin²A

.985 - .985 sin²A = .8118 sin²A

.985 = 1.7968 sin²A

sinA = .74

A = 47.73

B = 49.27

c / sin C = b / sin B

c = b sinC / sinB

= 44 x sin 83 / sin 49.27

= 44 x .9925 / .7578

= 57.62