Which of the following four statements is/are accurate with respect to glycolysis?

A. Glycolysis involves the conversion of monosaccharides into glucose.
B. Glycolysis involves the breakdown of glucose into pyruvate.
C. Glycolysis involves the conversion of pyruvate into glucose to glyceraldehyde-3-phosphate.
D. Glycolysis involves the conversion of pyruvate into acetyl CoA.

Explanation:

Meso compounds are optically inactive stereoisomers. Despite of having chiral carbon, they do not show  optical activity because it has a plane of symmetry in its structure itself. It is superposable on its mirror image.

Aldo sugars on reaction which reducing agents such as , NaBH₄ reduces the carbonyl group in the sugar to alcohol and gives corresponding alditol.

The D- aldohexose which on reduction gives a meso alditol are allose and galactose.

The structure is shown in the image below. Thus, in allitol and galactitol formed have internal plane of symmetry which makes the optically inactive.

The answer is in the photo

The molarity of the luminol stock solution is calculated by dividing the number of moles of luminol (0.08464 moles) by the volume of the solution in liters (0.075 L), resulting in a molarity of 1.13 M.

The ideal way to characterise stock solutions is as concentrated solutions with precise, known concentrations that will be diluted for later usage in laboratories. Although you have the option to forego preparing stock solutions, doing so can simplify your workflow and save you a significant amount of time and money. To solve the question we need to first determine the number of moles of luminol and then use the volume of the solution to find the molar concentration.

The molecular weight of luminol (C₈H₇N₃O₂) is approximately 177.16 g/mol. The number of moles of luminol can be calculated by dividing the mass of luminol by its molecular weight:

Number of moles of luminol = Mass of luminol / Molecular weight of luminol
Number of moles of luminol = 15.0 g / 177.16 g/mol = 0.08464 moles

Next, we convert the volume of water from milliliters to liters:

Volume in liters = 75.0 mL × (1 L / 1000 mL) = 0.075 L

Now we can calculate the molarity of the luminol solution:

Molarity (M) = Number of moles of solute / Volume of solution in liters
Molarity (M) = 0.08464 moles / 0.075 L

Molarity (M) = 1.13 M

The stock solution of luminol has a molarity of 1.13 moles per liter (1.13 M).

Answer : The entropy change of the system is, 361.83 J/K

Solution :

Formula used :

[tex]\Delta S=\frac{n\times \Delta H_{vap}}{T_b}[/tex]

where,

[tex]\Delta S[/tex] = entropy change of the system = ?

[tex]\Delta H[/tex] = enthalpy of vaporization = 34.6 kJ/mole

n = number of moles of diethyl ether = 4.20 mole

[tex]T_b[/tex] = normal boiling point = [tex]26.5^oC=273+26.5=307.6K[/tex]

Now put all the given values in the above formula, we get the entropy change of the system.

[tex]\Delta S=\frac{4.20mole\times (34.6KJ/mole)}{307.6K}=0.36183kJ/K=0.36183\times 1000=361.83J/K[/tex]

Therefore, the entropy change of the system is, 361.83 J/K

The answer is in the photo

Answer: [tex]10.17\times 10^{-23}g[/tex]

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given atoms}}{\text{Avogadro's number}}=\frac{1}{6.023\times 10^{23}}=0.16\times 10^{-23}atoms[/tex]

1 mole of zinc weighs = 65.38 g

[tex]0.16\times 10^{-23}[/tex]  moles of Zinc weigh = [tex]\frac{63.58}{1}\times 0.16\times 10^{-23}=10.17\times 10^{-23}g[/tex]

Thus mass of one atom of Zinc weigh [tex]10.17\times 10^{-23}g[/tex]

Answer:

weight percent of Iodine in the solution is 13.34%

Explanation:

Weight percentage -

weight percentage of A is given as , the weight of the substance A by weight of the total solution multiplied by 100.

i.e.

weight % A = weight of A / weight of solution * 100

From the question ,

weight of Iodine = 7.50 g

weight of carbon tetrachloride = 48.7 g

iodine is the solute and carbon tetrachloride is the solvent ,

Since,

solution = solute + solvent.

Hence,

weight of solution = weight of solute ( iodine ) + weight of solvent ( carbon tetrachloride)

weight of solution =  7.50 g + 48.7 g = 56.20 g

now,

weight of Iodine is calculated as,

weight % iodine = weight of iodine / weight of solution * 100

weight % iodine = 7.50 g / 56.20 g * 100

weight % iodine = 13.34 %

Answer:

Disorder

Explanation:

Entropy is a measure of disorder.

Entropy is a measure of disorder.

Hey there!:

Pentane is a 5 member hydrocarbon

1 chloropentane can be on either side and will still have thesame name because of the IUPAC rules .

To determine the relative amounts of products obtained from radical chlorination of an alkane, both probability (the number of hydrogens that can be abstracted that will lead to the formation of the particular product) and reactivity (the relative rate at which a particular hydrogen is abstracted) must be taken into account.

The precise ratios differ at different temperatures.

For 3- chloro pentane , There are two hydrogen only and the reactivity  is same for all  2° hydrogen

The 2°  hydrogen % should = 100-22 = 78% ( all remaining are 2° hydrogen only)

since all are equally recative 2° hydrogen  , proportionate distribution will be there.

% Yield  =   {( Total hydrogen which can give 3 choloropentane) /  (all hydrogen of 2° ) } * 78

             =  (2/6)* 78 =  26%  

Hope this helps!

In the chlorination of pentane, if 1-chloropentane is 22% of the mixture and secondary hydrogens are equally reactive as in monochlorination, the percentage of 3-chloropentane in the mixture is approximately 15%. The percentages are derived taking into consideration that the ratio of monochlorination among the different positions in pentane (primary, secondary, tertiary) is 3:2:0.

In organic chemistry, different isomers of a compound can be formed via a process of substitution like chlorination. In the chlorination of pentane, we have a mixture of isomers, including 1-chloropentane and 3-chloropentane, among others. The question stated that the given percentage of 1-chloropentane is 22%. If the reactivity of secondary hydrogens in pentane is equal to that of monochlorination, then it implies the formation of 2-chloropentane and 3-chloropentane have the same ratio because both have their Cl bound to a secondary carbon. Thus, the distribution of monochlorination among 3 positions (primary, secondary, and tertiary) is in a ratio 3:2:0 for 1,2 and 3-chloropentane respectively.

To derive the individual rates of formation, we'd have to divide each by the total, e.g., for 1-chloropentane (primary position), it would be 3/5 = 0.60 or 60%. However, the question gives us a value of 22% for the 1-chloropentane, which suggests our chlorination is only 22/60 = 0.37 or 37% effective. Applying this scale factor to 2,3 chloropentane (secondary position), we get 0.37*40 = 14.8%. Therefore, 3-chloropentane will constitute around 15% of the mixture.

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Hey there!:

Option: C) a solution that is 0.10 M HC₂H₃O₂and 0.10 M LiC₂H₃O₂

Reason : Buffer mixture means mixture of either weak acid and salt of weak acid or weak base and salt of weak base.

Here acetic acid is a weak acid and its Li salt

Hope that helps!

Answer:

PV=nRT

n = PV/RT

n = m/Mm

m/Mm = PV/RT

m = MmPV/RT

T in kelvin = T Celsius + 273.15 = 293.15 K

m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)

mass in grams = 82.8 grams  

Explanation:

Ideal gases formula is PV=nRT, where:

P is the pressure (1.39 atm in this case)

V is the volume (55.0 L in this case)

R is the gas constant (0.08206 L.atm/K.mole)

T is the temperature (20.0C) should be converted to Kelvin

all the unit should correspond to the one in the R.

we also know that to find the mass, we can use number mole with the formula number of mole(n) = mass (m) divided by the molar mass (Mm). therefore we substituted that in the formula and make (m) the subject of the formula.

we found the mass to be 82.8 grams

PV=nRT is the formula for ideal gases, where P is the pressure and R is the constant and T is the temperature (1.39 atm in this case)

[tex]m = MmPV/RT\\m = (26.04 x 1.39 x 55)/(0.08206 x 293.15)[/tex]

Thus, the mass was discovered to be 82.8 grams.

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Answer:

The student conclude that the sample of benzoic acid is impure.

Explanation:

The observed melting point of benzoic acid when a student melts his/her sample is low than the actual value. The reason for this might be:

(a) The most probable reason is that the sample is impure. Impurities in the sample leads to lowering the value of the melting point. The reason for the phenomenon is that when impurity is present in the compound, the pattern of the crystal lattice disturbs and thus it less amount of heat is require to break the lattice.

(b) There may be some experimental errors like:

The student conclude that the sample is impure.

Final answer:

The student's benzoic acid sample is likely impure, as indicated by a melting point range significantly lower than the expected 121-122 degrees Celsius for a pure sample.

Explanation:

If a good sample of benzoic acid typically melts at 121-122 degrees Celsius, but a student's sample melted over a range of 105-115 degrees Celsius, it could indicate that their benzoic acid sample is impure. The presence of impurities in a compound will generally lower the melting point and cause the substance to melt over a broader temperature range. This can be seen in a MelTemp apparatus where samples with varying amounts of impurities, such as those containing acetanilide, melt at lower temperatures and have a wider melting range compared to pure benzoic acid.

Answer:

Percent ionization of HCOOH is 3.69%

Explanation:

To calculate percent ionization of HCOOH, we have to construct an ICE table to determine changes in concentrations at equilibrium.

              [tex]HCOOH\rightleftharpoons HCOO^{-}+H^{+}[/tex]

    I:                0.125                                       0             0

    C:                -x                                           +x            +x

    E:              0.125-x                                     x              x

species inside third bracket represent equilibrium concentrations

So,      [tex]\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}=K_{a}(HCOOCH)[/tex]

   or,  [tex]\frac{x^{2}}{0.125-x}= 1.77\times 10^{-4}[/tex]

   or, [tex]x^{2}+0.000177x-0.0000221 = 0[/tex]

[tex]x=\frac{-0.000177+\sqrt{(0.000177)^{2}+(4\times 0.0000221)}}{2}[/tex]

or, [tex]x=4.61\times 10^{-3}[/tex] M

So, [tex][H^{+}]=4.61\times 10^{-3}M[/tex]

Percent ionization of formic acid  = [tex]\frac{[H^{+}]}{initial concentration of HCOOH}\times 100[/tex] = [tex]\frac{0.00461}{0.125}\times 100[/tex] = 3.69%

The percent ionization of the acid had been the concentration of hydrogen ion with respect to the acid concentration. The percent ionization of formic acid is 3.69%.

The acid dissociation constant has been the concentration of the acid dissociated into the constituent ions.

The dissociation of formic acid is given as:

[tex]\rm HCOOH\;\rightleftharpoons\;H^+\;HCOO^-[/tex]

The acid dissociation constant (Ka) for formic acid is given as:

[tex]\rm Ka=\dfrac{[H^+][HCOO^-]}{[HCOOH]}[/tex]

Substituting the concentration of the ions and the acid from the ICE table attached.

[tex]\rm 1.77\;\times\;10^{-4}=\dfrac{[x][x]}{[0.125-x]} \\\\x=4.61\;\times\;10^-3[/tex]

The hydrogen ion concentration in the solution has been 0.00461 M.

Substituting the concentration for the percent ionization of the formic acid:

[tex]\rm Percent\;ionization=\dfrac{[H^+]}{[HCOOH]}\;\times\;100\\\\ Percent\;ionization=\dfrac{0.00461}{0.125}\;\times\;100\\\\ Percent\;ionization=3.69\;\%[/tex]

The percent ionization of formic acid is 3.69%.

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Answer:

Must be either ether or ester.

Explanation:

The incorrect statement is that the monosaccharides must ether or esters.

The main and compulsory functional group in a monosaccharide are aldehyde and ketones.

The other statements are true

1) smallest monosaccharides are composed of three carbons: the smallest monosaccharide is glyceraldehyde.

a) aldoses and ketoses : they have either aldehyde group or ketone group

example: glucose is aldose and fructose if keotse

c) exist in many isomeric forms: there can be many arrangement on chiral carbons thus many stereoisomers are possible

e) Two monosaccharides can combine to form a disaccharide: yes for example glucose combines with fructose to give sucrose.

Answer: 0.35 moles

Explanation:

Avogadro's Law: This law states that volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

[tex]V\propto n[/tex]   (At constant temperature and pressure)  

[tex]\frac{V_1}{n_1}=\frac{V_2}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of gas = 52.0 L

[tex]V_2[/tex] = final volume of gas = 13.0 L

[tex]n_1[/tex] = initial number of moles = 1.40

[tex]n_2[/tex] = final number of moles = ?

Now put all the given values in the above equation, we get the final pressure of gas.

[tex]\frac{52.0}{1.40}=\frac{13.0}{n_2}[/tex]

[tex]n_2=0.35moles[/tex]

Therefore, the moles of the gas that will remain is 0.35 moles.

Hey there!:

convert mass of H2O to mol  , use given chemical equation to calculate moles of H2 produced , use ideal gas equation to calculate volume  

convert mass of H2O to mol  :

mol of H2O = mass / molar mass

=  15.0 / 18.0

= 0.833 mol

use given chemical equation to calculate moles of H2 produced

from given equation, 4 mol of H2 is formed from 4 mol of H2O

So,

moles of H2 = moles of H2O

= 0.833 mol

use ideal gas equation to calculate volume

use:

P*V = n*R*T

(745/760) atm * V = 0.833 mol * 0.0821 atm.L/mol.K * (20+273) K

0.9803 * V = 0.833 * 0.0821*293

V = 20.4 L

Answer: 20.4 L

Answer:

molar mass of nicotine will be 162.16g/mol

Explanation:

The mass of nicotine taken = 0.60g

The volume of solution = 12mL

the osmotic pressure of solution = 7.55 atm

Temperature in kelvin =298.15K (25+ 273.15)

The formula which relates osmotic pressure and concentration (moles per L) is:

π = MRT

Where

π = osmotic pressure (unit atm) = 7.55 atm

M = molarity (mol /L)

T= temperature = (K) = 298.15 K

R = gas constant = 0.0821 L atm /mol K

Putting values

[tex]7.55=MX0.0821X298.15[/tex]

Therefore

[tex]M=\frac{7.55}{0.0821X298.15}=0.308\frac{mol}{L}[/tex]

Molarity is moles of solute dissolve per litre of solution

The volume of solution in litre = 0.012 L

[tex]molarity=\frac{moles}{V}[/tex]

[tex]moles=molarityXvolume=0.308X0.012=0.0037mol[/tex]

we know that

[tex]moles=\frac{mass}{ymolarmass}[/tex]

molar mass = [tex]\frac{mass}{moles}=\frac{0.60}{0.0037}=162.16\frac{g}{mol}[/tex]

Using hydrogen and Lindlar catalyst the triple bond will be hydrogenated to a double one with a cis conformation.  

Answer: The density of gold in [tex]g/cm^3[/tex] is [tex]19.22g/cm^3[/tex]

Explanation:

Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

[tex]\text{Density}=\frac{\text{Mass of the object}}{\text{Volume of the object}}[/tex]

We are given:

Density of gold = [tex]1200lb/ft^3[/tex]

Using conversion factors:

1 lb = 453.6 g

1 feet = 12 inches

1 inch = 2.54 cm

Converting given quantity into [tex]g/cm^3[/tex], we get:

[tex]\Rightarrow (\frac{1200lb}{ft^3})\times (\frac{453.6g}{1lb})\times (\frac{1ft}{12inch})^3\times (\frac{1inch}{2.54cm})^3\\\\\Rightarrow 19.22g/cm^3[/tex]

Hence, the density of gold in [tex]g/cm^3[/tex] is [tex]19.22g/cm^3[/tex]

Answer:

79.0 g

Explanation:

1. Gather the information in one place.

MM:    148.89        253.81

           2NaI + Cl2 → I2 + 2NaCl

m/g:                         67.3

2. Moles of I2

n = 67.3 g × (1 mol/253.81 g) = 0.2652 mol I2

2. Moles of NaI needed

From the balanced equation, the molar ratio is 2 mol NaI: 1 mol I2

n = 0.028 76 mol I2× (2 mol NaI/1 mol I2) = 0.5303 mol NaI

3. Mass of NaI

m = 0.5303 mol × (148.89 g/1 mol) = 79.0 g NaI

It takes 79.0 g of NaI to produce 67.3 g of I2.

We need approximately 79.5 g of sodium iodide, NaI, to produce 67.3 g of iodine, I2, based on the stoichiometry of the reaction 2NaI(aq)+Cl2(g)⇾I2(s)+2NaCl(aq).

The question is about determining the amount of sodium iodide, NaI needed to produce a certain amount of iodine, I2, based on the equation 2NaI(aq)+Cl2(g)⟶I2(s)+2NaCl(aq). To solve this, we invoke stoichiometry. From the balanced equation, we know that 2 moles of NaI yield 1 mole of I2. Therefore, if we want to find the mass of NaI needed to produce 67.3 g of I2, we need to convert the mass of I2 to moles using its molar mass (about 253.8 g/mol), then multiply that by the ratio of moles of NaI to moles of I2 (which is 2:1), and then convert that to grams using the molar mass of NaI (about 149.9 g/mol). So, the calculation would be as follows: (67.3 g I2)/(253.8 g/mol) * 2 (moles of NaI/mole of I2) * 149.9 g/mol = about 79.5 g of NaI.

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Final answer:

The percent yield of the reaction producing tungsten and water is approximately 13.66%. To find this, the theoretical yield of tungsten was calculated using stoichiometry, and compared with the actual yield of water, assuming a 1:1 correspondence between tungsten and water produced by the reaction.

Explanation:

To calculate the percent yield of the reaction, we need to know the theoretical yield and the actual yield. The theoretical yield can be calculated using stoichiometry based on the balanced chemical equation for the reaction. In this case, we are given the production of water as a byproduct of the reaction between tungsten(VI) oxide (WO3) and hydrogen gas. We're also provided the mass of water produced (8.01 g, since the density of water is given as 1.00 g/mL and volume is 8.01 mL).

The balanced reaction is: WO3 + 3H2 → W + 3H2O

To find the theoretical yield of tungsten, we need the molar mass of WO3 and W:

Using the molar mass, we can calculate the moles of WO3 reacted:

74.1 g WO3 × (1 mol / 231.84 g) = 0.319 moles of WO3

Because the reaction stoichiometry is 1:1 for WO3 to W, the moles of tungsten produced should also be 0.319. Using the molar mass of tungsten, we can calculate the theoretical yield:

0.319 moles of W × 183.84 g/mol = 58.665 g of W

The actual yield given is 8.01 g of water, which we assume corresponds to the amount of tungsten produced since 3 moles of water are produced per mole of tungsten. Therefore, the percent yield is:

Percent Yield = (Actual Yield / Theoretical Yield) × 100%

Percent Yield = (8.01 g / 58.665 g) × 100% ≈ 13.66%

Answer:

1. [tex]60.0 mi/h=88 feet/seconds[/tex]

2. [tex]15 Ib/inch=267.86 kg/m[/tex]

3. [tex] 6.20 cm/hr=17,222.22 nm/s[/tex]

Explanation:

1. 60.0 mi/h to ft/s

1 mile = 5280 feet

1 hour = 3600 seconds

So, [tex]60.0 mi/h=\frac{60\times 5280 feet}{1\times 3600 s}=88 feet/seconds[/tex]

2. 15 Ib/in to kg/m

1 lb = 0.453592 kg

1 inch = 0.0254 m

So,[tex]15 Ib/inch=\frac{15\times 0.453592 kg}{1\times 0.0254 m}=267.86 kg/m[/tex]

3. 6.20 cm/hr to nm/s

1 cm = [tex]10^7 nm[/tex]

1 hour = 3600 seconds

So,[tex] 6.20 cm/hr=\frac{6.20\times 10^7 nm}{1\times 3600 s}=17,222.22 nm/s[/tex]

Answer: 0.11 M

Explanation:

The equation for the reaction is given as:

[tex]2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O[/tex]

According to the neutralization law,

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]HNO_3[/tex] solution = ?

[tex]V_1[/tex] = volume of [tex]HNO_3[/tex] solution = 0.105 L= 105 ml

[tex]M_2[/tex] = molarity of [tex]Ba(OH)_2[/tex] solution = 0.2 M

[tex]V_2[/tex] = volume of [tex]Ba(OH)_2[/tex] solution = 29.1 ml

[tex]n_1[/tex] = valency of [tex]HNO_3[/tex] = 1

[tex]n_2[/tex] = valency of [tex]Ba(OH)_2[/tex] = 2

[tex]1\times M_1\times 105=2\times 0.2\times 29.1[/tex]

[tex]M_1=0.11M[/tex]

Therefore, the concentration of [tex]HNO_3[/tex] will be 0.11 M.

Answer: The chemical equations are given below.

Explanation:

A balanced chemical reaction follows law of conservation of mass.

This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.  

Single displacement reaction is defined as the reaction in which more reactive metal displaces a less reactive metal from its chemical reaction.

General equation representing single displacement reaction follows:

[tex]AB+C\rightarrow CB+A[/tex]

C is more reactive element than element A.

The reactivity of metals is judged by the series known as reactivity series. Elements lying above in the series are more reactive than the elements lying below in the series.

Lead lies below in the reactivity series than iron. Thus, it will not replace iron from its chemical reaction.

[tex]Pb(s)+Fe(NO_3)_2(aq.)\rightarrow \text{No reaction}[/tex]

Iron lies above in the reactivity series than lead. Thus, it will easily replace lead from its chemical reaction.

[tex]Fe(s)+Pb(NO_3)_2(aq.)\rightarrow Fe(NO_3)_2(aq.)+Pb(s)[/tex]

Hence, the chemical equations are given above.

In the first scenario, no reaction occurs when solid lead is placed into a solution of Fe(NO₃)₂ because lead is less reactive than iron. In the second scenario, a reaction occurs when iron is placed into a solution of Pb(NO₃)₂ because iron is more reactive than lead. The resulting balanced chemical equation is: Fe(s) + Pb(NO₃)₂(aq) -> Fe(NO₃)₂(aq) + Pb(s).

In chemistry, a chemical reaction occurs when two or more atoms bond together to form molecules or when bonded atoms break apart. We refer to the substances used in the beginning of a reaction as the reactants and the substances found at the end of the reaction as the products.

The first scenario involves a strip of solid lead (Pb) placed into a solution of Fe(NO₃)₂. No reaction happens in this case because lead is less reactive than iron. In the second scenario, a strip of iron (Fe) is placed into a solution of Pb(NO₃)₂. A reaction occurs because iron is more reactive than lead. The balanced chemical equation for this reaction is: Fe(s) + Pb(NO₃)₂(aq) -> Fe(NO₃)₂(aq) + Pb(s).

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Answer : The correct pairs of elements likely to form ionic compounds are, potassium and sulfur, magnesium and chlorine.

Explanation :

Ionic compound : These are the compounds in which the atoms are bonded through the ionic bond. The ionic bond are formed by the complete transfer of the electrons.

That means, the atom which looses the electron is considered as an electropositive atom and the atom which gains the electron is considered as an electronegative atom.

And these bonds are formed between one metal and one non-metal.

From the given pairs of elements, potassium and sulfur, magnesium and chlorine are form ionic compound because potassium and magnesium are the metals and sulfur and chlorine are the non-metals. So, they can easily form ionic compound by the complete transfer of electrons.

While the other pairs, sodium and potassium are the metals, nitrogen and iodine, chlorine and bromine, helium and oxygen are the non-metals. They do not form ionic bond.

Hence, the correct pairs of elements likely to form ionic compounds are, potassium and sulfur, magnesium and chlorine.

Sodium and potassium, potassium and sulfur, and magnesium and chlorine are likely to form ionic compounds due to large differences in electronegativity.

Ionic compounds are formed when one element transfers electrons to another element, resulting in the formation of positive and negative ions. Elements with different electronegativities are more likely to form ionic compounds. Sodium and potassium, potassium and sulfur, and magnesium and chlorine are likely to form ionic compounds because of the large difference in electronegativity between them. In contrast, nitrogen and iodine, chlorine and bromine, helium and oxygen have similar electronegativities and are more likely to form covalent compounds.

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Answer:

see explanation...

Explanation:

The central nitrogen atom in the azide ion has a formal charge of +1. The other nitrogen atoms have a formal charge of -1.

The resonance structure of azide ion and hydrazoic acid is shown in the image attached to this answer. We can see that the central nitrogen atom in the azide ion has a formal charge of +1. The other nitrogen atoms have a formal charge of -1.

In hydrazoic acid, hydrogen is attached to one of the end nitrogen atoms while the other nitrogen atoms have a formal of +1 and -1 respectively.

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Answer: True

Explanation: Intensive property does not depend on the amount of matter. Specific internal energy is an intensive property.

Hey there!:

Intensive property , does not depend upon the amount  of matter , specific internal energy is an intensive property , due to word specific  . TRUE

Hope this helps!

Answer:

146,575 grams of ethylene glycol would need to be dissolved in 15 kg of pure water

The boiling point of the solution is 181.95°C.

Explanation:

Mass of ethylene glycol = x

Molar mass of ethylene glycol = 2 × 12 g/mol + 2 × 32 g/mol+ 4 × 1= 62g/mol

Moles of ethylene glycol =[tex]\frac{x}{62 g/mol}[/tex]

Mass of solvent that is water = 15 kg

The molal freezing point depression constant for water [tex]K_f=1.86 K kg/mol[/tex]

Molality of the solution:

[tex]Molality=\frac{\text{Moles of compound}}{\text{Mass of solvent (kg)}}[/tex]

[tex]Molality=m=\frac{x}{62 g/mol\times 15 kg}[/tex]

Depression in freezing point of water =[tex]\Delta T_f=20^oC=293.15 K[/tex]

((T)°C =T+ 273.15 K)

[tex]\Delta T_f=K_f\times m[/tex]

[tex]293.15 K=1.86 K kg/mol\times \frac{x}{62 g/mol\times 15 kg}[/tex]

x = 146,575 g

Boiling point of this solution =[tex]T_b[/tex]

The molal boiling point elevation constant for water[tex]K_b = 0.52 K kg/mol[/tex]

[tex]\Delta T_b=K_b\times m[/tex]

[tex]\Delta T_b=0.52 K kg/mol\times \frac{146,575 g}{62 g/mol\times 15 kg}[/tex]

[tex]\Delta T_b=81.95K[/tex]

Normal boiling point of water is = T = 373.15 K

[tex]\Delta T_b=T_b-T[/tex]

[tex]T_b=\Delta T_b+T=81.95K+373.15 K=455.1 K=181.95^oC[/tex]

The boiling point of the solution is 181.95°C.

Answer:

Is a dietary essential, especially for athletes shuttles .

Explanation:

Carnitine is a dietary essential, especially for athletes shuttles .

Carnitine shuttles fatty acids from the cytosol into the mitochondria, where they are broken down for energy in a process called oxidation. It is a nutrient associated with lipid metabolism and is involved in the Carnitine Shuttle, but it is not an essential dietary element for everyone, particularly those with already sufficient levels.

Carnitine is a nutrient involved in lipid metabolism within the mitochondria, especially for the transportation of fatty acids. In cellular metabolism, carnitine is not a dietary essential, but it shuttles fatty acids from the cytosol into the mitochondria, enabling their breakdown for energy production. This process, known as the Carnitine Shuttle, involves the conversion of fatty acyl-CoA molecules into acyl-carnitine by the enzyme carnitine palmitoyltransferase I, which allows the molecule to be transported across the inner mitochondrial membrane. Once within the matrix, carnitine acyltransferase II converts acyl-carnitine back into acyl-CoA for oxidation.

Despite being marketed as a supplement for athletes and individuals looking to enhance fat catabolism, carnitine supplementation has not been universally agreed upon as effective for those with sufficient carnitine levels. The role of carnitine in facilitating fatty acid oxidation is crucial, but it does not involve the shuttling of oxaloacetate from the mitochondria to the cytosol.

Answer:

Mole fraction of glycerol is 0.02666.

Volume of the mixture is 99.75 mL.

Explanation:

Volume of glycerol,V = 10 mL

Mass of glycerol = m

Density of glycerol =[tex]\rho = 1.25802 g/cm^3=1.25802 g/mL[/tex]

[tex]m=\rho \times V=1.25802 g/mL\times 10 mL=12.5802 g[/tex]

Volume of water,V' = 90 mL

Mass of water= m'

Density of water =[tex]\rho '= 0.99708 g/cm^3=0.99708 g/mL[/tex]

[tex]m'=\rho '\times V'=0.99708 g/mL\times 90 mL=89.7372 g[/tex]

Mole fraction of glycerol =[tex]\chi_g=\frac{n_g}{n_g+n_w}[/tex]

[tex]\chi_g=\frac{\frac{12.5802 g}{92.09 g/mol}}{\frac{12.5802 g}{92.09 g/mol}+\frac{89.7372 g}{18 g/mol}}=0.02666[/tex]

Volume of the solution: v

Mass of the solution = M = 12.5802 g + 89.7372 g =102.3174 g[/tex]

Density of the mixture = [tex]\rho _{mix}=1.02567 g/cm^3=1.02567 g/mL[/tex]

[tex]v=\frac{M}{\rho _{mix}}=\frac{102.3174 g}{1.02567 g/mL}=99.75 mL=99.75 cm^3[/tex]

Volume of the mixture is 99.75 mL.

Theoretically the volume of the solution should be 100 ml.But experimentally the volume of the solution is 99.75 ml which is less than the theoretical volume.

This is because water molecules and glycerol molecules are getting associated with each other due to hydrogen bonding which results in less volume of the mixture.

Answer:

28.12 g Na2CrO4

Explanation:

To solve this question, we have to work with stoichiometry:

Then, if we have 45.5 g of Silver Nitrate:

[tex]45,5 g AgNo3 x \frac{1 mol AgNo3}{169.87 g AgNo3} = 0.2678 mol AgNo3\\[/tex]

We know that for each Na2CrO4's mol, we need 2 AgNo3's moles, so, we can calculate the amount of AgNo3's moles as follows:

[tex]0.2678 mol AgNo3 * \frac{1 mol Na2CrO4}{2 Mol AgNO3} =0.134 mol Na2CrO4\\[/tex]

and, now we can calculate the grams of sodium chromate as:

[tex]0.134 mol Na2CrO4 *\frac{209.9714 g Na2CrO4}{ molNa2CrO4} =28.12 g Na2CrO4[/tex]

To react completely with 45.5 g of silver nitrate, you would need approximately 21.6 g of sodium chromate based on the stoichiometric ratio from the balanced chemical equation.

In order to determine how many grams of sodium chromate is needed to react completely with the silver nitrate, we first look at the balanced chemical equation given: 2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3. This equation tells us that it takes 2 moles of AgNO3 to react completely with 1 mole of Na2CrO4. Now, we need to know the molar mass of AgNO3 and Na2CrO4 which are approximately 170 g/mol and 162 g/mol respectively.

Then, calculate the number of moles of AgNO3 we have by dividing the mass by the molar mass: 45.5 g / 170 g/mol = 0.267 moles. According to the balanced chemical equation, 0.267 moles of AgNO3 will react completely with half the amount of Na2CrO4. So, the number of moles of Na2CrO4 = 0.267 moles / 2 = 0.1335 moles. To find how many grams this is, we multiply by the molar mass of Na2CrO4: 0.1335 moles * 162 g/mol = 21.63 g. Therefore, you would need approximately 21.6 g of sodium chromate to react completely with 45.5 g of silver nitrate.

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