Is ribose a reducing or non reducing sugar?

Answers

Answer 1
Ribose is used in RNA and deoxyribose is used in DNA. The deoxy- designation refers to the lack of an alcohol, -OH, group as will be shown in detail further down. Ribose and deoxyribose are classified as monosaccharides, aldoses, pentoses, and are reducing sugars.
CARBOHYDRATES: Di, poly-Carbohydrates
Classification: Glucose
Carbo - Isomers: Galactose
Answer 2

Ribose is a reducing sugar. A reducing sugar is a carbohydrate that can undergo a redox reaction, in which it donates electrons to another chemical species.

This is usually observed when the sugar opens its ring structure to form an aldehyde or ketone functional group.

Ribose, a five-carbon sugar, can form an open-chain structure with an aldehyde functional group. In this form, it can donate electrons and act as a reducing agent in certain chemical reactions, such as the reduction of other compounds like Benedict's reagent during laboratory tests for reducing sugars.

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Related Questions

what are the angles a and b in the actual molecule of which this is a lewis structure note for advanced students give the ideal angles and don t worry about small differences from the ideal that might be caused by the fact that different electron groups may have slightly different sizes

Answers

Answer:

The answer is "120 C  and 109.5 C".

Explanation:

The carbon atom is hybridized by sp2. This angle of connection thus is 120 degrees. Alkene, specifically both carbons which are in the C=C, are an instance of carbon with sp2 hybridized atom's nucleus. Those three hybridized orbits were linked to certain other atoms forming sigma connections. Its remaining 2p orbital makes a pi link with 2p orbit by the side-overlap of all the other carbon. O is hybridized inside the [-OH] Group. The optimal bond angle therefore is [tex]109.5^{\circ}[/tex].

[tex]a= 120 \ C\\\\b= 109.5 \ C[/tex]

Water put into a freezer compartment in the same refrigerator goes into a state of less molecular disorder when it freezes. Is this an exception to the entropy principle

Answers

Answer:

No it is not an exception to this principle

Explanation:

Work was carried out by this compressor to reduce the entropy of ice. What this means is that the ice gave out heat which is as a result  of the work that the compressor was putting in. there are violations of this principle

the entropy principle has that the entropy of the universe is always going to be more than 0 (system + surrounding). in this question, the that of the system is negative while that of the surrounding is positive.  As the refrigerator was cooling the water, the air outside was getting heated.  Outside this refrigerator, the gain in entropy is more than the entropy that was lost in the water.

the entropy of the universe once again is more than 0.

1. Most of the chemicals included in your General Chemistry Lab kit can be discarded down a drain. Describe a situation in which you would need to neutralize a chemical before discarding down a drain.

Answers

Answer: Chemicals like acids and bases are harmful and must be neutralized before draining.

Explanation:

A strong acid or strong base is required to be diluted or neutralized before it is discarded in the drain as if is discarded without diluting and neutralization it can spill and splash from sink or drain and can harm people in chemistry lab, moreover the fumes of the discarded chemical on spilling can cause respiratory tract burning and can even cause fire hazard so it must be converted into less harmful form and then must be drained.  

Please help me order these bonds urgent

Answers

Answer:

From least polar covalent to most polar covalent;

S-I< Br-Cl < N-H< Te-O

From most ionic to least ionic

Cs-F> Sr-Cl> Li- N> Al-O

Explanation:

Electro negativity refers to the ability of an atom in a bond to attract the shared electrons of the bond towards itself.

Electro negativity difference between two atoms is a key player in the nature of bond that exists between any two atoms. A large difference in electron negativity leads to an ionic bond while an intermediate difference in electro negativity leads to a polar covalent bond.

Based on electro negativity differences, the bonds in the answer have been arranged in order of increasing polar covalent nature or decreasing ionic nature.

What enzyme below is an exoenzyme?
A. Casease
B. Citrase
C. Catalase
D. Oxidase

Answers

Casease! Good luck!

Elements that have the same number of electron rings are ?

Answers

Answer:

are in the same orbital

Explanation:

Answer:

are in the same orbit

Explanation:

A chemical reaction in a bomb calorimeter evolves 3.86 kJ of energy in the form of heat. If the temperature of the bomb calorimeter increases by 4.17 K, what is the heat capacity of the calorimeter?

Answers

Answer:

925.66 J/K

Explanation:

Applying,

Q = CΔt............. Equation 1

Where Q = amount of heat, C = heat capacity of the calorimeter, Δt = rise in temperature.

make C the subject of the equation

C = Q/Δt.............. Equation 2

From the question,

Given: Q = 3.86 kJ = 3860 J, Δt = 4.17K

Substitute into equation 2

C = 3860/4.17

C = 925.66 J/K

What is the fourth quantum number of the 3p electron in aluminum,
1s^2 2s^2 2p^6 3s^2 3p^1?
A. ms = -1
B. ms = +1/2
C. ms=0
D. ms = +1​

Answers

Explanation:

here's the answer to your question

The quantum numbers are defined as the set of four numbers with the help of which we can get complete information about the electrons in an atom. The fourth quantum number is the spin quantum number. Here ms for 3p electron in 'Al' is ms = + 1/2. The correct option is B.

The quantum number which describes the spin orientation of the electron is defined as the spin quantum number. Since the electron can spin only in two ways, clockwise and anti-clockwise, the spin quantum number can have either the value +1/2 or -1/2 depending upon the direction of spin.

Thus for 3p electron in 'Al' ,ms is option B.

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For each of the following compounds, indicate the pH at which 50% of the compound will be in a form that possesses a charge and at which pH more than 99% of the compound will be in a form that possesses a charge.

ClCH2COOH (pKa = 2.86)
CH3CH2NH+3 (pKa = 10.7)

Express your answer using two decimal places

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.
b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.
c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.
d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Answers

Answer:

a. 2..86 b. 4.86 c. 10.7 d. 8.7

Explanation:

a. Determine a pH at which 50% of ClCH2COOH will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base (or charged form) and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 2.86

b. Determine a pH at which pH more than 99% of ClCH2COOH will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.99x while the acidic concentration is remaining 1 % (1 - 0.99)x = 0.01x

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base (or charged form) = 0.99x and [HA] = concentration of acid = 0.01x.

pH = pKa + log0.99x/0.01x

pH = pKa + log0.99/0.01

pH = 2.86 + log99

pH = 2.86 + 1.996

pH = 4.856

pH ≅ 4.86

c. Determine a pH at which 50% of CH3CH2NH+3 will be in a form that possesses a charge.

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA]

where [A⁻] = concentration of conjugate base and [HA] = concentration of acid.

At 50% concentration, [A⁻] = [HA] ⇒ [A⁻]/[HA] = 1

So, pH = pKa + log[A⁻]/[HA]

pH = pKa + log1

pH = pKa = 10.7

d. Determine a pH at which pH more than 99% of CH3CH2NH+3 will be in a form that possesses a charge.

Let x be the concentration of the acid. Since 99% of it should possess a charge, the basic concentration is 0.01x while the acidic concentration is remaining 99 % (1 - 0.01)x = 0.99x (which possesses the charge).

Using the Henderson-Hasselbalch equation,

pH = pKa + log[A⁻]/[HA] where [A⁻] = concentration of conjugate base = 0.01x and [HA] = concentration of acid = 0.99x.

pH = pKa + log0.01x/0.99x

pH = pKa + log1/99

pH = 10.7 - log99

pH = 10.7 - 1.996

pH = 8.704

pH ≅ 8.7

Solid aluminum (AI) and oxygen (0) gas react to form solid aluminum oxide (AIO). Suppose you have 7.0 mol of Al and 13.0 mol of o, in a reactor. Suppose as much as possible of the Al reacts. How much will be left? Round your answer to the nearest 0.1 mol mol 0.

Answers

Answer:

[tex]n_{O_2}^{leftover}=7.7mol[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set up the corresponding chemical equation:

[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]

In such a way, we calculate the moles of aluminum consumed by 13.0 moles of oxygen in the reaction, by applying the 4:3 mole ratio between them:

[tex]n_{Al}=13.0molO_2*\frac{4molAl}{3molO_2} =17.3molAl[/tex]

This means that Al is actually the limiting reactant and oxygen is in excess, for that reason we calculate the moles of oxygen consumed by 7.0 moles of aluminum:

[tex]n_{O_2}=7.0molAl*\frac{3molO_2}{4molAl} =5.3molO_2[/tex]

Thus, the leftover of oxygen is:

[tex]n_{O_2}^{leftover}=13.0mol-5.3mol\\\\n_{O_2}^{leftover}=7.7mol[/tex]

Whereas all the aluminum is assumed to be consumed.

Regards!

A covalent bond is formed by the following process

Answers

Answer:

Covalent bonding occurs when pairs of electrons are shared by atoms.

Explanation:

Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability.

The homework question reads:
"A sample of gas in a cylinder of volume 3.42 L at 298 K
and 2.57 atm expands to 7.39 L by two different pathways.
Path A is an isothermal, reversible expansion. Path B has two
steps. In the fi rst step, the gas is cooled at constant volume to
1.19 atm. In the second step, the gas is heated and allowed to
expand against a constant external pressure of 1.19 atm until
the final volume is 7.39 L. Calculate the work for each path.

Answers

Answer:

Explanation:

this guy on brainly already did it:

Alleei

Virtuoso

4.8K answers

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Answer :  The work done for path A and path B is -685.3 J and -478.1 J  respectively.

Explanation :

To calculate the work done for path A :

First we have to calculate the moles of the gas.

where,

= initial pressure of gas  = 2.57 atm

= initial volume of gas  = 3.42 L

n = moles of gas  = ?

R = gas constant = 0.0821 atm.L/mol.K

T = temperature of gas  = 298 K

Now put all the given values in the above formula, we get:

According to the question, this is the case of isothermal reversible expansion of gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

The expression used for work done will be,

where,

w = work done on the system = ?

n = number of moles of gas  = 0.359 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 298 K

= initial volume of gas  = 3.42 L

= final volume of gas  = 7.39 L

Now put all the given values in the above formula, we get :

Thus, the work done of path A is, -685.3 J

To calculate the work done for path B :

The formula used for isothermally irreversible expansion is :

where,

w = work done

= external pressure = 1.19 atm

= initial volume of gas = 3.42 L

= final volume of gas = 7.39 L

Now put all the given values in the above formula, we get :

Thus, the work done of path B is, -478.1 J

5.60g of glyceraldehydes was dissolved in 10ml of a solvent and placed in a 50mm cell if the rotation is 1.74 calculate the specific rotation?​

Answers

Answer:

6.214 degrees-mL/gdm

Explanation:

The specific rotation α' = α/LC where α = observed rotation, L = length of tube and C = concentration of solution.

Given that α = 1.74, L = length of cell = 50 mm = 0.50 dm and C = m/V where m = mass of glyceraldehyde = 5.60 g and V = volume = 10 ml

So, C = m/V = 5.60 g/10 ml = 0.560 g/ml

Since α' = α/LC

substituting the values of the variables into the equation, we have

α' = α/LC

α' = 1.74/(0.50 dm × 0.560 g/ml)

α' = 1.74/(0.28 gdm/l)

α' = 0.006214 °mL/gdm

α' = 6.214 °mL/gdm

α' = 6.214 degrees-mL/gdm

how many of the electrons in a molecule of ethane are not involved in bondind​

Answers

Ethane consists of 6C−H bonds and 1C−C bond. Total number of bonds is 7. Each bond is made up of two electrons

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The decomposition of ethyl amine, C2H5NH2, occurs according to the reaction: C2H5NH2(g)⟶C2H4(g)+NH3(g) At 85∘C, the rate constant for the reaction is 2.5 x 10-1 s-1. What is the half-life (in sec) of this reaction?

Answers

Answer:

2.772 seconds

Explanation:

Given that;

t1/2 = 0.693/k

Where;

t1/2 = half life of the reaction

k= rate constant

Note that decomposition is a first order reaction since the rate of reaction depends on the concentration of one reactant

t1/2 = 0.693/2.5 x 10-1 s-1

t1/2= 2.772 seconds

A sample of oxygen occupies 1.00 L. If the temperature remains constant, and the pressure on the oxygen is decreased to one third the original pressure, what is the new volume

Answers

Answer:

3.00 L

Explanation:

P₁V₁ = P₂V₂

V₁ = 1.00 L

P₁ = (x) atm

P₂ = [tex]\frac{1}{3}[/tex] · (P₁) = [tex]\frac{x}{3}[/tex]

V₂ = unknown

(x atm)(1.00 L) = ( [tex]\frac{x}{3}[/tex] atm)(V₂)

divide both sides by ( [tex]\frac{x}{3}[/tex] atm)

( 1.00x )( [tex]\frac{3}{x}[/tex] ) = V₂

x cancels out

(1.00)(3) = V₂

V₂ = 3.00 L

Use the Ka values for weak acids to identify the best components for preparing buffer solutions with the given pH values.

Name Formula Ka
Phosphoric acid H3PO4 7.5 x 10^-3
Acetic acid CH3COOH 1.8 x 10^-5
Formic acid HCOOH 1.8 x 10^-4

pH 1.9 =_________
pH 5.0 = ________
pH 3.9= ________

Answers

Answer:

pH= 1.9 then [tex]H_{3} PO_{4}[/tex]

pH = 5.0 , [tex]CH_{3} COOH[/tex]

pH = 3.9 , HCOOH

As we know range left [tex]pH= pKa+/- 1[/tex]

The compound IF5 contains Question 16 options: polar covalent bonds with partial negative charges on the F atoms. ionic bonds. polar covalent bonds with partial negative charges on the I atoms. nonpolar covalent bonds.

Answers

Answer:

See explanation

Explanation:

The molecule IF5 possesses five I-F polar bonds. However, the presence of polar bonds does not automatically imply that the molecule will be polar.

The geometry of the molecule is very important in determining the polarity of a compound. Since IF5 has a lone pair of electrons, the molecule is bent and as such there is a permanent dipole moment created in the molecule thereby making IF5 polar in nature.

For the following list of acids, rank the acids in strength from weakest acid to strongest acid.

a. FCH2OH
b. F2CHOH
c. CH3OH
d. F3COH

Answers

Answer:

CH3OH < FCH2OH < F2CHOH < F3COH

Explanation:

Let us recall that, for a carboxylic acid, the dissociation of the acid yields;

RCOOH ⇄RCOO^- + H^+

The ease of dissociation and release of the hydrogen ion depends on the nature of the group designated R.

When R is is a highly electronegative element, the -I inductive effect causes the hydrogen to become less tightly held by the C-Cl bond.

As the number of electron withdrawing substituents increaseses, the acid ionizes much more and becomes stronger.CH3OH < FCH2OH < F2CHOH < F3COH

Hence, the order of decreasing acid strength is;

Enzyme catalyzing breakdown of atp to adp

Answers

Answer:

ATP hydrolase

Explanation:

Enzymes are biological catalysts which perform diverse functions in the body. Enzymes are specific in their mode of action because an enzyme fits into its substrate as a key fits into a lock.

The particular enzyme that catalyzes the breakdown of ATP to ADP is ATP hydrolase. The phosphate released by the action of this enzyme is used in the phosphorylation of other compounds thereby making them more reactive.

Each set of quantum numbers to the correct sub shell description

Answers

The image of the chart that I attach will allow you to answer the question since you did not include in your question of the quantum numbers. I hope this helps :)

Classify the processes as endothermic or exothermic.

a. Ice melting
b. Water condensing on surface
c. Baking a cake
d. The chemical reaction inside an instant cold pack.
e. A car using gasoline

Answers

endothermic absorbs heat

exothermic gives heat

a. endothermic

b. exothermic

c. endothermic

d. exothermic

a. Ice melting - endothermic

b. Water condensing on the surface - exothermic

c. Baking a cake - endothermic

d. The chemical reaction inside an instant cold pack - endothermic

e. A car using gasoline - exothermic

What is an exothermic and endothermic reaction?

An exothermic reaction can be described as a thermodynamic chemical reaction that emits energy from the system to its surroundings usually in the form of light, heat, or sound.

While an endothermic reaction can be described as an opposite of an exothermic reaction where the energy gains in the form of heat. In exothermic chemical reactions, the bond energy is transformed into thermal energy.

In exothermic reactions, the reaction happens the form of the kinetic energy of molecules when the energy is released. The release of energy is due to the electronic transition of electrons from one energy level to another.

The burning of gasoline, and water condensation is also an exothermic reaction in which energy is released while ice melting and baking cake is an endothermic reaction.

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4) In water, Vanillin, C8H8O3, has a solubility of 0.070 moles of vanillin per liter of solution at 25C. What will be produced if 5.00 g of vanillin are added to 1 L of water at 25 C

Answers

Answer:

An unsaturated solution.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly realize we need to calculate the grams of vanillin in 0.070 moles by using its molar mass of 152.15 g/mol:

[tex]m=0.070mol*\frac{152.15 g}{1mol} =10.65g[/tex]

Thus, since the solubility is 10.65 g per 1 L of solution, we can notice 5.00 g will complete dissolve and produce an unsaturated solution.

Best regards!

Determine the molecular formula of a compound if it is composed of 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen. The molar mass is 176.12 g/mol.
A) C3H8O3.
B) CH2O.
C) C2H3O2.
D) C3H4O3.

Answers

Answer:

No correct answer listed. See explanation for defense.

Explanation:

Given

C: 40.92% => 40.92g/100wt => (40.92/12)moles C = 3.41 moles O

H:   4.58% =>   4.58g/100wt =>      (4.58/1)moles H = 4.58 moles H

O: 54.50% => 54.5g/100wt =>     (54.5/16)moles O = 3.41 moles O

Empirical ratio => C : H: O => (3.41/3.41) : (4.58/3.41) : (3.41/3.41) => 1 : 1.34 : 1

=> C : H : O => 3(1 : 1.34 : 1) => 3 : 4 : 3 => Empirical Formula C₃H₄O₃

Molecular Weight = Empirical Formula Wt x N

         176.12          =           88  x  N

N = whole number multiple of empirical formula = 176.12/88 = 2

∴ Molecular Formula => (C₃H₄O₃)₂ => C₆H₈O₆

Note => Only ionic compounds (salts) have subscripts reduced to lowest whole number ratios. Molecular compounds as C₆H₈O₆ are not reduced to lowest whole number ratios. Therefore, there is no correct answer in the answer choice list for the 'Molecular Formula'. Doc :-)

Hazmat products warnings or labels allowed in fc

Answers

Answer:

The Hazmat products warnings or labels allowed in fc include:

1. Fully Regulated Aerosol Placard

2. Fully Regulated Flammable Solid Placard  

3. Fully Regulated Flammable  

4. Lithium-Ion/Metal Battery label

Explanation:

Hazmat products (including explosives, flammable liquids and solids, and gases, etc.) are classified as dangerous substances and materials that pose a risk to people during their storage, handling, or transportation.  The requirement for this Hazmat classification is to show that the identified products require diligence, carefulness, and alertness in handling, transporting, and storing them.  The reason for this is that mishaps can occur.  Some of them can also cause fire outbreaks.

Calculate the mass of sodium phosphate in aqueous solution to fully react with 37 g of chromium nitrate(III) an aqueous solution?(report answer in grams and only three Sigg figs do not put the unit)

Answers

Answer:

41 g

Explanation:

The equation of the reaction is;

Cr(NO3)3(aq)+Na3PO4(aq)=3NaNO3(s)+CrPO4(aq)

Number of moles of chromium nitrate = 37g/ 146.97 g/mol = 0.25 moles

1 mole of sodium phosphate reacts with 1 mole of chromium nitrate

x moles of sodium phosphate react as with 0.25 moles of chromium nitrate

x= 1 × 0.25/1

x= 0.25 moles

Mass of sodium phosphate = 0.25 moles × 163.94 g/mol

Mass of sodium phosphate = 41 g

Write a balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution. Be sure to add physical state symbols where appropriate.

Answers

Answer:

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

Explanation:

An oxidation reaction reaction refers to a reaction in which electrons are lost. In this case, we are about to see the full balanced half-reaction for the oxidation of liquid water to aqueous hydrogen peroxide in basic aqueous solution.

The full equation is;

O2(g) + 2H2O(l) --------> 2H2O2(aq) + 2e

So, two electrons were lost in the process.

Sometimes in lab we collect the gas formed by a chemical reaction over water . This makes it easy to isolate and measure the amount of gas produced.
Suppose the CO, gas evolved by a certain chemical reaction taking place at 50.0°C is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 132. mL. Calculate the mass of CO, that is in the collection tube. Round your answer to 2 significant digits.

Answers

Answer:

0.17 g

Explanation:

Since the volume of gas collected is 132 mL, we need to find the number of moles of gas present in 132 mL.

So, number of moles, n = volume of gas, v/molar volume, V

n = v/V where v = 132 mL = 0.132 L and V = 22.4 L

So, substituting the values of the variables into the equation, we have

n = v/V

n = 0.132 L/22.4 L

n = 0.005893 mol

We then need to calculate the molar mass of CO, M = atomic mass of carbon + atomic mass of oxygen = 12 g/mol + 16 g/mol = 28 g/mol

Also, number of moles of gas, n = m/M where m = mass of CO and M = molar mass of CO

m = nM

m = 0.005893 mol × 28 g/mol

m = 0.165004 g

m ≅ 0.17 g to 2 significant digits

42 Organic compound may have names ending in -ane, -ene, -ol or -oic acid. How many of these endings indicate the compounds contain double bonds in their molecules? * (1 Point)​

Answers

Answer: Organic compounds ending with the name (-ene) indicate that the compounds contain double bonds in their molecules.

Explanation:

Organic compounds are those molecules that contains carbon atoms (as their main element), hydrogen and oxygen which are usually present. The presence of numerous organic compounds is due to the following properties of carbon:

--> the exceptional ability of carbon atoms to catenate, that is, to combine with one another to form straight chains, branched chains or ring compounds containing many carbon atoms.

--> The ease with which carbon combines with hydrogen, oxygen, Nitrogen and halogens

--> The ability of carbon atoms to form single, DOUBLE or triple bonds.

The organic compound that has the name ending with -ene are known as the alkenes. The members of the alkene series are formed from the alkanes by the removal of two hydrogen atoms and the introduction of a DOUBLE BOND in the carbon chain. They are named after the corresponding alkanes by changing the -ane ending to -ene.

Note: the systematic name of a compound is formed from the root hydrocarbon by adding a suffix and prefixes to denote the substitution of the hydrogen atoms.

How many molecules are in
6.0 moles of methane (CH4)?

Answers

Answer:

[tex]{ \tt{1 \: mole = 6.02 \times {10}^{23} \: molecules }} \\ { \tt{6.0 \: moles = (6 \times 6.02 \times {10}^{23}) \: molecules }} \\ = { \bf{3.612 \times {10}^{24} \: molecules}}[/tex]

Other Questions
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