15. Ammonium nitrate, NH4NO3, and aluminum powder react explosively producing nitrogen gas, water vapor and aluminum oxide. Write the balanced equation and calculate the enthalpy change for this reaction.

Answer:

[tex]m_{MgO}=0.239gMgO[/tex]

Explanation:

Hello,

In this case, the chemical reaction between magnesium and oxygen to yield magnesium oxide is:

[tex]2Mg+O_2\rightarrow 2MgO[/tex]

In such a way, for 0.144 g of magnesium reacting with sufficient oxygen, the mass of magnesium oxide, whose molar mass is 40.3 g/mol (2:2 mole ratio) turns out:

[tex]m_{MgO}=0.144gMg*\frac{1molMg}{24.3gMg} *\frac{2molMgO}{2molMg}* \frac{40.3gMgO}{1molMgO} \\\\m_{MgO}=0.239gMgO[/tex]

Best regards.

Answer:

C3H7OH → C3H6 + H20

Explanation:

If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.

Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.

Answer:

B and C. Just finished my lesson on Edge.

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:

It is an example of elimination reaction through the E2 mechanism.

Explanation:

The reaction between 1-iodobutane and pyridine is an example of an E2 (bimolecular elimination) elimination reaction.

Pyridine acts predominantly as a base and the given alkyl halide is a primary alkyl halide. Both of these two factors facilitate the E2 mechanism.

Here, both H and Cl are eliminated in a single step to produce 1-butene as the product of the reaction.

The reaction mechanism and the structure of the product are shown below.

The mechanism by which 1-iodobutane reacts with pyridine is by the E2 mechanism.

The E2 mechanism process (Bimolecular Elimination) is a one-step reaction mechanism whereby carbon-hydrogen (C-H) and carbon-halogen (C-X) bonds split to generate a double bond. (C = C πbond).

The following characteristics of the E2 reaction are:

From the information given:

Pyridine functions primarily as a base, and the alkyl iodide in question is a primary alkyl halide that helps in the E2 mechanism.

In this case, both H and Cl are removed in a single step, yielding 1-butene as the byproduct of the reaction.

Learn more about the Bimolecular Elimination reaction here:

https://brainly.com/question/13607621

Answer: stay away from doors and windows.

Explanation:

to aviod geting hit by glass

Answer:

Stay away from doors and windows.

Explanation:

Always stay in the center of the room during a tornado storm. Avoid windows, doors, and corners. If you’re near a window, the glass can shatter and hurt you.

Answer:

Kc = 166.7

[Fe³⁺] =  0.18 M

[SCN⁻] = 2×10⁻⁴ M

Explanation:

In the reaction of Fe³⁺ and SCN⁻, it is formed a complex:

Fe³⁺  +  SCN⁻  ⇄  FeSCN²⁺             Kc

Let's make the expression for Kc →  [FeSCN²⁺] / [Fe³⁺] . [SCN⁻]

5.7×10⁻⁵ / 9.5×10⁻⁴. 3.6×10⁻⁴  = 166.7

We determine the mmoles, we add from each reactant:

18 ml . 0.2M = 3.6 mmoles of Fe³⁺

2 ml . 0.002M = 4×10⁻³ mmoles of SCN⁻

General form of the dilution equation is:

Concentrated [C] . Concentrated Volume = Diluted [C] . Diluted Volume

Total volume = 20mL

[Fe³⁺]: 3.6 mmoles /20mL = 0.18 M

[SCN⁻]: 4×10⁻³ mmoles /20mL = 2×10⁻⁴ M

The value should be 1.67 x 10^2

The initial concentration should be 0.18 M and 2.0 x 10^(-4) M

The value is

= 5.7 x 10^(-5)/(9.5 x 10^(-4) x 3.6 x 10^(-4))

= 167

= 1.67 x 10^2

we know that

Initial moles  = volume x concentration

So,

= 18/1000 x 0.200

= 0.0036 mol

Now

Initial moles  = volume x concentration

= 2/1000 x 0.0020

= 4.0 x 10^(-6) mol

So,

Total volume should be

= 18 + 2

= 20 mL

= 0.02 L

Now

Initial concentration   

= moles /total volume

= 0.0036/0.02

= 0.18 M

Now

Initial concentration

= moles  /total volume

= 4.0 x 10^(-6)/0.02

= 2.0 x 10^(-4) M

Learn more about equation here: https://brainly.com/question/4083100

Answer:

See explanation

Explanation:

In this case, we have to keep in mind the valence electrons for each atom:

N => 5 electrons

O => 6 electrons

If the formula is [tex]N_2O_4[/tex], we will have in total:

[tex](5*2)+(6*4)=34~electrons[/tex]

Additionally, we have to remember that each atom must have 8 electrons. So, for oxygens 5 and 3 we will have 3 lone pairs and 1 bond (in total 8 electrons. For oxygens, 6 and 4 we will have 2 lone pairs and 2 bonds (in total 8 electrons) and for nitrogens 1 and 2 we will have 4 bonds (in total 8 electrons).

To find the hybridization, we have to count the atoms and the lone pairs around the nitrogen. We have 3 atoms and zero lone pairs. If we take into account the following rules:

[tex]Sp^3~=~4[/tex]

[tex]Sp^2~=~3[/tex]

[tex]Sp~=~2[/tex]

With this in mind, the hybridization of nitrogen is [tex]Sp^2[/tex].

See figure 1

I hope it helps!

The central nitrogen atoms in N2O4 are both sp2 hybridized.

The Lewis structure shows the number of electron pairs that surround the atoms in a molecule as dots. It is quite easy to determine the number of valence electrons in a molecule simply by observing its Lewis dot structure.

The molecule N2O4 has 34 valence electrons as shown in its dot electron structure. The central nitrogen atoms in N2O4 are both sp2 hybridized as shown. The formal charges on each atom in N2O4 are also shown.

Learn more:

Answer:

The answer to your question is given below.

Explanation:

Aluminium has atomic number of 13. Thus, the electronic configuration of aluminium can be written as:

Al (13) —› 1s² 2s²2p⁶ 3s²3p¹

The orbital diagram is shown on the attached photo.

Answer: screen shot

Explanation:

Answer:

3.383x10⁻³ micromoles of HgCl

Explanation:

The chemist adds 170mL of a 1.99x10⁻⁵mmol/L Mercury (I) chloride, HgCl.

The solution contains 1.99x10⁻⁵milimoles of HgCl in 1L. That means in 170mL = 0.170L there are:

0.170L × (1.99x10⁻⁵milimoles HgCl / L) = 3.383x10⁻⁶ milimoles of HgCl.

Now, in 1milimole you have 1000 micromoles. That means in 3.383x10⁻⁶ milimoles of HgCl you have:

3.383x10⁻⁶ milimoles of HgCl ₓ (1000micromoles / 1milimole) =

Answer:

[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].

One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.

Explanation:

Start by finding the formula for each of the compound.

Therefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between

The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].

Write down the equation using these chemical formulas.

[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.

Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:

[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].

Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.

There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.

Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.

In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.

One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].

Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.

One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].

Hence:

[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:

[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].

That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:

[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].

Answer:

The possible solution is to balance the rate of water removal from the well to the rate of natural recharge of the well from its underground aquifer.

Explanation:

A well is an excavation in the earth, made with the aim of extracting water from the aquifers. The water from a well can be drawn up by the means of a pump, containers, such as buckets, or by hand. Aquifers can also be recharged through a well.

Well draw down occurs when water from the well is drained faster than it is naturally recharged from the aquifer. This can be as a result of over pumping, extended drought, among other factors. The use of the high-powered motor in this case, for pumping, might be the possible cause of the well drying up. The situation might have resulted from the pump drawing out water from the well at a rate tat exceeds the rate at which it is recharged naturally, causing the well water to start drying up. There's also a possibility that the well is pumped indiscriminately, possibly leading to wastage of water.

The solution to this problem is to give the well a time duration for it to recharge itself. Then, the rate of recharges should be calculated and determined by an hydrologist. When all these is done, a pump with a motor power that does not exceed the calculated recharge rate should be used in place of the high-powered motor. Also, water usage should be brought to the minimum level to prevent unnecessary pumping due to excessive, wasteful use of water.

Answer:

[tex]\Delta G_{rxn}=42.7\frac{kJ}{mol}[/tex]

Explanation:

In this case, for the dissociation of hypochlorous acid, we know that the acid dissociation constant (Ka) is 2.9x10⁻⁸, which is related with the Gibbs free energy as shown below:

[tex]\Delta G_{rxn}=-RTln(K)[/tex]

But in this case K is just Ka, therefore, at 296 K, it turns out:

[tex]\Delta G_{rxn}=-8.314\frac{J}{mol*K}*296K*ln(2.9x10^{-8})\\\\\Delta G_{rxn}=42.7\frac{kJ}{mol}[/tex]

Such result, means that the reaction is nonspontaneous at the given temperature, it means it is not favorable (not easily occurring).

Best regards.

Answer:

the answer is in the diagram

Explanation:

when 2,4-dimethylpent-2-ene undergo electrophilic addition reaction in the presence of HBr to form 2-bromo-2,4-dimethylpentane, it firstly lead to an intermediate carbocation

A carbocation can be describe as an organic molecule, which serves as an intermediate, that has a carbon atom bearing a positive charge and three bonds instead of four

Answer:

CH3CH=NH2+>CH3CH2NH3+

Explanation:

A close examination of both structures will reveal that they are both amines hence they must have the polar N-H bond.

Electrons usually move towards the nitrogen atom and this makes both compounds acidic. We must also remember that some features of a compound may make it more acidic than another of close resemblance. Being more acidic may imply that the proton of the N-H is more easily lost.

CH3CH=NH2+ has an sp2 hybridized carbon atom in its structure which is known to be very electronegative due to increasing s character of the bond. It will withdraw electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ in comparison to CH3CH2NH3+

Answer:

a. Strong acid, pH = 1.69

b. Strong base, pH = 10

c. Strong base, pH = 11

d. Weak acid, pH = 4.90

e. Strong base, pH ≅ 7 (pH should be higher than 7, but the base is so diluted)

f. Weak base, pH = 10.96

g. Acidic salt, pH = 5.12

h. Basic salt, pH = 8.38

i. Neutral salt, pH = 7

j. Acidic salt, pH < 7

Explanation:

a. HBr →  H⁺  +  Br⁻

Hydrobromic acid is a strong acid.

pH = - log [H⁺]

- log 0.02 = 1.69

b. NaOH → Na⁺  +  OH⁻

Sodium hydroxide is a strong base.

pH = 14 - pOH

pOH = - log [OH⁻]

pH = 14 - (-log 0.0001) = 10

c. Ba(OH)₂ → Ba²⁺  +  2OH⁻

Barium hydroxide is a strong base

[OH⁻] = 2 . 0.0015 = 0.003M

pH = 14 - (-log 0.003) = 11

d. HCN + H₂O ⇄  H₃O⁺  + CN⁻

This is a weak acid, it reacts in water to make an equilibrium between the given protons and cyanide anion.

To calculate the [H₃O⁺] we must apply, the Ka

Ka = [H₃O⁺] . [CN⁻] / [HCN]

6.2×10⁻¹⁰ = x² / 0.25-x

As Ka is really small, we can not consider the x in the divisor, so we avoid the quadratic formula.

[H₃O⁺] = √(6.2×10⁻¹⁰ . 0.25) = 1.24×10⁻⁵

-log 1.24×10⁻⁵ = 4.90 → pH

e.  KOH →  K⁺  +  OH⁻

2×10⁻¹⁰ M → It is a very diluted concentration, so we must consider the OH⁻ which are given, by water.

In this case, we propose the mass and charges balances equations.

Analytic concentration of base = 2×10⁻¹⁰ M = K⁺

[OH⁻] = K⁺ + H⁺ → Charges balance

The solution's hydroxides are given by water and the strong base.

Remember that Kw = H⁺ . OH⁻, so H⁺ = Kw/OH⁻

[OH⁻] = K⁺ + Kw/OH⁻. Let's solve the quadratic equation.

[OH⁻] = 2×10⁻¹⁰ + 1×10⁻¹⁴ /OH⁻

OH⁻² = 2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴

2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴ - OH⁻²

We finally arrived at the answer [OH⁻] = 1.001ₓ10⁻⁷

pH = 14 - (- log1.001ₓ10⁻⁷) = 7

The strong base is soo diluted, that water makes the pH be a neutral value.

Be careful, if you determine the [OH⁻] as - log 2×10⁻¹⁰, because you will obtain as pOH 9.69, so the pH would be 4.31. It is not possible, KOH is a strong base and 4.30 is an acid pH.

f. Ammonia is a weak base.

NH₃ +  H₂O  ⇄  NH₄⁺  + OH⁻

Kb = OH⁻  .  NH₄⁺  /  NH₃

1.74×10⁻⁵ = x² / 0.05 - x

We can avoid the x from the divisor, so:

[OH⁻] = √(1.74×10⁻⁵ . 0.05) = 9.32×10⁻⁴

pH = 14 - (-log 9.32×10⁻⁴ ) = 10.96

g. NH₄Cl, an acid salt. We dissociate the compound:

NH₄Cl →  NH₄⁺  +  Cl⁻.  We analyse the ions:

Cl⁻ does not make hydrolisis to water. In the opposide, the ammonium can react given OH⁻ to medium, that's why the salt is acid, and pH sould be lower than 7

NH₄⁺  +  H₂O  ⇄  NH₃  +  H₃O⁺   Ka

Ka = NH₃  .   H₃O⁺ / NH₄⁺

5.70×10⁻¹⁰ = x² / 0.1 -x

[H₃O⁺] = √ (5.70×10⁻¹⁰  . 0.1) = 7.55×10⁻⁶

pH = - log 7.55×10⁻⁶ = 5.12

As Ka is so small, we avoid the x from the divisor.

h. CaF₂  →  Ca²⁺  +  2F⁻

This is a basic salt.

The Ca²⁺ does not react to water. F⁻ can make hydrolisis because, the anion is the strong conjugate base, of a weak acid.

F⁻  +  H₂O  ⇄  HF  +  OH⁻          Kb

Kb = x² / 2 . 0.2 - x

Remember that, in the original salt we have an stoichiometry of 1:2, so 1 mol of calcium flouride may have 2 moles of flourides.

As Kb is small, we avoid the x, so:

[OH⁻] = √(1.47×10⁻¹¹ . 2 . 0.2) = 2.42×10⁻⁵

14 - (-log 2.42×10⁻⁵) = pH → 8.38

i . Neutral salt

BaNO₃₂  →   Ba²⁺  +  2NO₃⁻

Ba²⁺ comes from a strong base, so it is the conjugate weak acid and it does not react to water. The same situation to the nitrate anion. (The conjugate weak base, from a strong acid, HNO₃)

pH = 7

j.  Al(NO₃)₃, this is an acid salt.

Al(NO₃)₃  →  Al³⁺  +  3NO₃⁻

The nitrate anion is the conjugate weak base, from a strong acid, HNO₃ so it does not make hydrolisis. The Al³⁺ comes from the Al(OH)₃ which is an amphoterous compound (it can react as an acid or a base) but the cation has an acidic power.

Al·(H₂O)₆³⁺  + H₂O ⇄  Al·(H₂O)₅(OH)²⁺  + H₃O⁺

Answer:

Fluorine, Chlorine, Bromine, or Iodine

Explanation:

These all have an ALMOST full valence shell. And they need one more electron so they'd react with hydrogen

Answer:

its chlorine

Explanation:

just trust me do i look like i would lie too you ;-)

btw i just took the test :-)

Answer:

= 0.2 mL.

Explanation:

Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:

[tex]C_{1} V_{1} = C_{2} V_{2}[/tex]

where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.

Let C1 = 0.5M, V2 = ?

C2 = 0.010M; V2 = 10mL

⇒Volume of stock solution to be diluted, V2

= [tex]\frac{10}{0.5}[/tex] × 0.010

= 0.2 mL.

Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.

I hope this was a detailed explanation given the missing details of "Trial 1" in the question.

Answer:

2 moles of HNO3

Explanation:

The equation seems to be balanced correctly. The problem is we done know what you started with. We will assume it is 3 moles of NO2.

If that is the case then 2 moles of HNO3 will be produced.

Answer:

Explanation:

Calcium bicarbonate  dissolved in hard water can easily be removed by heating the hard water . On heating , it decomposes to give calcium carbonate which is insoluble and therefore can be filtered out .

Ca( HCO₃)₂  =  CaCO₃ + CO₂ + H₂O.

In this way hardness of water is removed .

Answer:

The voltage is  [tex]E_{cell} = 0.944 \ V[/tex]

Explanation:

Generally the half reaction for Zn, Zn2 half-cell is mathematically represented as

    [tex]Zn_{(s)}[/tex]   ⇔   [tex]Zn^{2+}_{ (aq)} + 2e^-[/tex]    (reference study academy)

and the electric potential for this is a constant value

     [tex]E_{zn } = -0.7618 \ V[/tex]

Generally the half reaction for Al, Al3 half-cell is mathematically represented as

      [tex]Al^{3+} _{(aq)} + 3e^-[/tex]  ⇔  [tex]Al_{(s)}[/tex]

and the electric potential for this is constant value  

       [tex]E_{Al } = -1.662 \ V[/tex]

Therefore the cell potential for an electrochemical cell  is mathematically represented as

          [tex]E_{cell} = E_{zn } - E_{Al }[/tex]

substituting values

         [tex]E_{cell} = -0.718 - (-1.662)[/tex]

        [tex]E_{cell} = 0.944 \ V[/tex]

Answer:

A 50:50 mixture of ethylene glycol and water is effective both summer and winter extremes in temperature because of high boiling point of 106°C and low freezing point of about -37°C. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.

Explanation:

Ethylene glycol or antifreeze is an organic compound which is used in automobile engines as a coolant and also as an anti-freezing agent, however it does not conduct heat effectively as water due to its lower heat capacity. It has a freezing point of -12.9°C and boiling point of 197.3°C.

Water is also used as a coolant in automobile engine but it has a limited range due to its boiling point of 100°C. It is also not a good anti-freezing agent due to it high freezing point of 0.°C

However, when ethylene glycol is mixed with water in a ratio of  50:50, the property of the mixture is enhanced to both serve as a coolant and as an antifreeze. The boiling point is elevated to about 106°C while its freezing point is lowered to about -37°C.

This temperature range is effective for both summer and winter temperatures. In summer when the average daily temperature rises to about 22°c, the mixture will be effective in keeping the the automobile engine cool. Also in winter, when the average temperature falls below 0°C, the mixture will be effective as an antifreeze as it remains a liquid well below 0°C.

The molar mass of stearic acid is 18*AC+36*AH+2*AO=18*12+36*1+16*2=284g/mol.

n=m/M=2.831/284=0.01 moles

C18H36O2+27O2-->18CO2+18H2O

we have 18*0.01=0.18 moles of CO2

18*0.01=0.18 moles of H2O

0.01*948=9.48kJ from stearic acid

0.18*393.5=70.83kJ from CO2

0.18*241.826=43.52kJ from H2O

9.48+70.83+43.52=123.83kJ

123.83*4.184=518.10kcal

Answer:

The answer is 6 because the p sublevel holds 3 orbitals and since each orbital can hold 2 electrons, the answer is 3 * 2 = 6.

Answer:

6 electrons

Explanation:

Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. So the answer is 6 electrons.

A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll

Answer:

Catalyst

Explanation:

For the reaction:

[tex]Cl_2_(_g_)~+~3F_2_(_g_)->2ClF_3_(_g_)[/tex]

We have a main observation: When platinum is added the reaction goes faster. With this in mind, we have to remember the kinetic equilibrium theory. In figure 1, we have an energy diagram. In which we have an specific energy for the reagents and the products. When the reaction takes place, the reaction has to must go through an energy peak. This energy peak is called "activation energy". When platinum is added the activation energy decreases and the reaction can go faster. Therefore, platinum is a "catalyst", a substance with the ability to reduce the activation energy.

I hope it helps!

Answer:

a. New alpha- 1,6 linkages can only form if the branch has a free reducing end

b. The number of sites for enzyme action on a glycogen molecule is increased through alpha- 1,6 linkages

c. At least four glucose residues separate alpha-1,6 linkages

e. The reaction that forms alpha-1,6 linkages is catalyzed by a branching enzyme.

Explanation:

Glycogen i is the main storage polysaccharide in animals. It a homoplymer of (alpha-1-->4)-linked subunits of glucose molecules, with alpha-1--->6)-linked branches.

The alpha-1,6 branches are formed by the glycogen-branching enzyme which catalyzes the transfer of about 7 glucose residues from the non-reducing end of a glycogen branch having at least 11 residues to the C-6 hydroxyl group of a glucose residue which lies inside the same glycogen chain or another glycogen chain, thereby forming a new branch. This ensures that there are at least four glucose residues separating alpha-1,6 linkages.

The effect of branching is that it makes the glycogen molecule more soluble and also increases the number of non-reducing ends, thereby increasing the number of sites for the action of the enzymes glycogen phosphorylase and glycogen synthase.

Answer:

sodium

Explanation:

(Na) atom is paramagnetic and sodium is a na atom.

The cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.

FACILITATED DIFFUSION:

Learn more at: https://brainly.com/question/18122054?referrer=searchResults

Answer:

D- poor conductor

Explanation:

metallic properties decrease as we go on the right of the periodic table. Carbon is a non metal hence it is dull and a poor conductor.

it has a low density and is ductile.

Answer: Poor conductor

Explanation:

Answer:

6

Explanation:

Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6