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kupik [55]
3 years ago
12

The compound responsible for the characteristic smell of garlic is allicin, The mass of 1.00 mol of allicin, rounded to the near

est integer, is __________ g.
Chemistry
1 answer:
never [62]3 years ago
3 0

Answer:

The mass of 1.00 mol of allicin, rounded to the nearest integer, is 162 g

Explanation:

Allicin is a compound that derivates from the alliin which is produced by the catalyzis of an enzime.

The molecular formula is: C₆H₁₀OS₂.

Let's determine the molar mass which is the mass that corresponds to 1 mol

Molar mass C . 6 + Molar mass H . 10 + Molar mass O + Molar mass S . 2 =

12 g/mol . 6 + 1 g/mol . 10 + 16 g/mol + 32.06 g/mol . 2 =  162.1 g/mol

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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
3 years ago
True or False: As the ball rises from point 1 to point 3, it slows down
quester [9]
True, it would slow down.
7 0
3 years ago
The number, 6.022 x 1023 is known as a ___.
Gennadij [26K]

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The electron config for 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p3​
nadya68 [22]

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3 0
3 years ago
alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decima
svetlana [45]

A 1.0-L buffer solution contains 0.100 mol  HC2H3O2 and 0.100 mol  NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.

Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

Answer:

The pH of this solution = 5.06  

Explanation:

Given that:

number of moles of CH3COOH = 0.100 mol

volume of the buffer solution = 1.0 L

number of moles of NaC2H3O2 = 0.100 mol

The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.

we know that concentration in mole = Molarity/volume

Then concentration of [CH3COOH] = \mathtt{ \dfrac{0.100  \ mol}{ 1.0  \  L }}  = 0.10 M

The chemical equation for this reaction is :

\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}

The conjugate base is CH3COO⁻

The concentration of the conjugate base [CH3COO⁻] is  = \mathtt{ \dfrac{0.100  \ mol}{ 1.0  \  L }}  

= 0.10 M

where the pka (acid dissociation constant)for CH3COOH = 4.74

If 0.035 mol of NaOH is added  to the original buffer, the concentration of NaOH added will be = \mathtt{ \dfrac{0.035  \ mol}{ 1.0  \  L }} = 0.035 M

The ICE Table for the above reaction can be constructed as follows:

                  \mathtt{CH_3COOH \ \ \   +  \ \ \ \ OH^-  \ \ \to \ \ CH_3COO^-  \ \ \ + \ \ \  H_2O}

Initial             0.10               0.035         0.10                  -

Change        -0.035          -0.035       + 0.035              -

Equilibrium    0.065              0              0.135               -

By using  Henderson-Hasselbalch equation:

The pH of this solution = pKa + log \mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}

The pH of this solution = 4.74 + log \mathtt{\dfrac{0.135}{0.065}}

The pH of this solution = 4.74 + log (2.076923077 )

The pH of this solution = 4.74 + 0.3174

The pH of this solution = 5.0574

The pH of this solution = 5.06    to two decimal places

7 0
3 years ago
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