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Basile [38]
2 years ago
8

Is bread a molecule or a compound?

Chemistry
2 answers:
Anna007 [38]2 years ago
4 0

Answer:

Molecule

Explanation:

This is what I found

The chemical building blocks of bread are proteins and starch. Starch molecules are long, chained polymers of simple sugars (such as glucose) joined end to end by chemical bonds. Proteins on the other hand are more complex, made up of varying combinations of different amino acids.

Morgarella [4.7K]2 years ago
3 0

it's molecule.......

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which elements are likely to have a larger atomic radius than silicon (Si).aluminum (Al)carbon (C)sulfur (S)tin (Sn)
dimulka [17.4K]
D.) Tin (Sn) are likely to have a larger atomic radius than silicon....
7 0
3 years ago
A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t
jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x  (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is  initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for  weak acids for the equilibria:

protonated lidocaine + H₂O   ⇆  lidocaine + H₃O⁺

                      protonated lidocaine          lidocaine        H₃O⁺

Initial(M)               0.22                                       0                  0

Change                   -x                                      +x                 +x

Equilibrium          0.22 - x                                  x                    x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] =  x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸  = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22  - x  ≈ 0.22

and solve for x. The pH  will then  be the negative log of this value.

8.71 x 10⁻⁷  = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0
3 years ago
Which of the following reactions is the least energetic? Question 18 options: ATP + H2O → ADP + Pi ATP + H2O → AMP + PPi AMP + H
Mama L [17]

Answer:

The correct answer is AMP+H2O→ Adenosine + pi

Explanation:

The above reaction is least energetic because there is no phosphoanhydride bond present with adenosine mono phosphate.Phospho anhydride bond is an energy rich bond.

As a result hydrolysis of AMP generates very little amount of energy in comparison to the hydrolysis of ATP and ADP.

   

8 0
3 years ago
What is the name of the ionic compound Csmc010-1.jpgS? cesium sulfur cesium sulfide dicesium sulfur dicesium sulfide
nika2105 [10]
Answer is: cesium sulfide.
Formula for ionic compound is: Cs₂S.
Ionic compound <span>electrically </span>neutral, cesium has charge +1 (2· (+1) = +2) and sulfur has charge -2.
Cesium sulfide is whide powder, soluble in water, pyrophoric in dry air, reductant.
5 0
3 years ago
Read 2 more answers
A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO
Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL

Putting values in above equation, we get:

2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL

M_2=0.336M

Thus, the concentration of NaOH is, 0.336 M

3 0
3 years ago
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