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tiny-mole [99]
2 years ago
11

compound of aspartame is a dipeptide that is often used as a sugar substitute which functional groups are present

Chemistry
1 answer:
Scrat [10]2 years ago
5 0

Answer:

Carboxyl, primary amine, amide, ester, and phenyl.

Explanation:

The functional groups present in the compound of aspartame are carboxyl, primary amine, amide, ester, and phenyl. Aspartame is an artificial non-saccharide sweetener which is 200 times sweeter than sucrose. This aspartame is commonly used as a sugar substitute in many foods and beverages. It has the trade names such as NutraSweet, Equal, and Canderel.

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Which are the hybrid orbitals of the carbon atoms in the following molecules? (a) H3C―CH3 sp sp2 sp3 (b) H3C―CH═CH2 sp sp2 sp3 (
kherson [118]

Answer:

(a)  sp³    sp³

   H₃<u>C</u> - <u>C</u>H₃

(b)     sp³           sp²

     H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂

                sp²

(c)     sp³        sp    

    H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH

              sp         sp³

(d)     sp³    sp²    

       H₃<u>C</u> - <u>C</u>H=O

Explanation:

Alkanes or the carbons with all the single bonds are sp³ hybridized.

Alkenes or the carbons with double bond(s) are sp² hybridized.

Alkynes or the carbons with triple bond are sp hybridized.

Considering:

(a) H₃C-CH₃ , Both the carbons are bonded by single bond so both the carbons are sp³ hybridized.

Hence,

 sp³    sp³

H₃<u>C</u> - <u>C</u>H₃

(b) H₃C-CH=CH₂ , The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp² hybridized because they are bonded by double bond.

Hence,

   sp³           sp²

H₃<u>C</u> - <u>C</u>H = <u>C</u>H₂

         sp²

(c) H₃C-C≡C-CH₂OH , The carbons of the methyl group and alcoholic group are sp³ hybridized as it is boned via single bonds. The rest 2 carbons are sp hybridized because they are bonded by triple bond.

Hence,

   sp³        sp    

H₃<u>C</u> - <u>C</u> ≡ <u>C</u> - <u>C</u>H₂OH

         sp         sp³

(d)CH₃CH=O, The carbon of the methyl group is sp³ hybridized as it is boned via single bonds. The other carbon is sp² hybridized because it is  bonded by double bond to oxygen.

Hence,

   sp³    sp²    

 H₃<u>C</u> - <u>C</u>H=O

     

8 0
3 years ago
Could someone help with this? Much appreciated!
alekssr [168]

Answer:

The 3rd answer down.

Na²O (sodium oxide) will be a base when exposed to water H²O

Explanation:

Sodium Oxide Na²O, will become Sodium Hydroxide after being exposed to water (at 80% I believe).

The oxygen ion in Na²O has 2 extra electrons which makes it highly charged and very attractive to hydrogen ions. The attraction is so strong that when Na²O comes in contact with H²O, the O(-2) strips off a hydrogen from water, forming 2 x OH ions which of course are still strongly basic.

6 0
2 years ago
At what temperature (°C) will a 10.00 g sample of neon gas exert a pressure of 96.7 kPa in a
Eva8 [605]

Answer : The temperature of neon gas will be, 221.0 K

Explanation :

To calculate the temperature of neon gas we are using ideal gas equation.

PV=nRT\\\\PV=\frac{w}{M}RT

where,

P = pressure of neon gas = 96.7 kPa = 0.955 atm

Conversion used : (1 atm = 101.3 kPa)

V = volume of neon gas = 9.50 L

T = temperature of neon gas = ?

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

w = mass of neon gas = 10.00 g

M = molar mass of neon gas = 20 g/mole

Now put all the given values in the ideal gas equation, we get:

(0.955atm)\times (9.50L)=\frac{10.00g}{20g/mole}\times (0.0821L.atm/mole.K)\times T

T=221.0K

Therefore, the temperature of neon gas will be, 221.0 K

7 0
2 years ago
Be sure to answer all parts. A 0.365−mol sample of HX is dissolved in enough H2O to form 835.0 mL of solution. If the pH of the
Marta_Voda [28]

Answer:

The Ka is 9.11 *10^-8

Explanation:

<u>Step 1: </u>Data given

Moles of HX = 0.365

Volume of the solution = 835.0 mL = 0.835 L

pH of the solution = 3.70

<u>Step 2:</u> Calculate molarity of HX

Molarity HX = moles HX / volume solution

Molarity HX = 0.365 mol / 0.835 L

Molarity HX = 0.437 M

<u />

<u>Step 3:</u> ICE-chart

[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4

Initial concentration of HX = 0.437 M

Initial concentration of X- and H3O+ = 0M

Since the mole ratio is 1:1; there will react x M

The concentration at the equilibrium is:

[HX] = (0.437 - x)M

[X-] = x M

[H3O+] = 1.995*10^-4 M

Since 0+x = 1.995*10^-4   ⇒ x=1.995*10^-4

[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M

[X-] = x = 1.995*10^-4 M

<u>Step 4: </u>Calculate Ka

Ka = [X-]*[H3O+] / [HX]

Ka = ((1.995*10^-4)²)/ 0.437

Ka = 9.11 *10^-8

The Ka is 9.11 *10^-8

4 0
3 years ago
What physical structure has spines that protect the plant from predators?
Bumek [7]
Flower !! !!! !! Yea yes yes
6 0
2 years ago
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