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aleksandrvk [35]
3 years ago
11

At the midpoint of a titration curve ________. At the midpoint of a titration curve ________. The concentration of a conjugate b

ase is twice that of the concentration of a conjugate acid The pH equals the pKa The ability of the solution to buffer is at its least effective The concentration of a conjugate base is 1/2 that of the concentration of a conjugate acid
Chemistry
1 answer:
VARVARA [1.3K]3 years ago
8 0

Answer:

The concentration of a conjugate base is twice that of the concentration of a conjugate acid

Explanation:

This is because at the midpoint, the number of moles of acid equals half the number of moles of base at the midpoint. This means that at the midpoint, half the analyte has been titrated.

Since the concentration of the conjugate acid is half that of the conjugate base at the midpoint, this implies that the concentration of the conjugate base is twice that of the conjugate acid.

<u>Thus, at the midpoint of a titration curve the concentration of a conjugate base is twice that of the concentration of a conjugate acid </u>

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Kimchi is traditional Korean food made by fermenting cabbage or other vegetables in a mixture of spices and salt. Kimchi origina
Yuki888 [10]

Answer:

C. Water passes into the salt solution, dehydrating bacterial cells and making them harmless.

Explanation:

The salt solution is hypertonic to the bacterial cells and as such, water molecules will move from the bacterial cells into the salt solution, dehydrating the cells and rendering them harmless.

Option A is also true but it is irrelevant to the question asked. Option B and D are wrong.

The correct option is C.

3 0
3 years ago
12
VARVARA [1.3K]

Answer:

B

Explanation:

the number of protons never changes even if the atomic mass changes

5 0
3 years ago
A plot of 1/[BrO-] vs. time is linear and the slope is equal to 0.056 M-1s-1. If the initial concentration of BrO- is 0.65 M, ho
finlep [7]

Answer:

time taken for one-half of the BrO⁻ ion to react is t= 27.45 secs

Explanation:

equation of reaction

3BrO⁻(aq) → BrO₃⁻(aq) + 2Br⁻(aq) (second order reaction)

given

the slope of the graph is 0.056M⁻¹s⁻¹ = k(constant)

initial concentration [A]₀ = 0.65M

for second order reaction,we can calculate the time taken for one-half of the BrO- ion to react using:

\frac{1}{[A]} =\frac{1}{[A]}₀ ⁺ k × t

where initial concentration [A]₀ = 0.65M

[A] = [A]₀÷2 = 0.325M

\frac{1}{0.325M} = \frac{1}{0.65M} + 0.056M⁻¹s⁻¹ × t

3.077= 1.54 + 0.056t

3.077-1.54=0.056t

1.537=0.056t

t= 27.45 secs

4 0
3 years ago
Does the length of the carbon chain affect the boiling point of alkanes? If so, how? provide evidence to support your answer? (b
kirill115 [55]
<h2>(a) </h2>

Yes. B.p. tend to increase with the number of carbon atoms in the backbone.

<h3>Explanation</h3>

Straight-chain alkanes have the general molecular formula C_nH_{2n+2}. Adding carbon atoms to the chain increases the number of electrons in each molecules.

Alkane molecules are not polar without any functional groups. The only possible force between them is London Dispersion Force (a.k.a. induced dipole). Electrons shift within molecules to create instantaneous dipole in this type of force.

Molecules with a large number of electrons experience the most significant shift. London Dispersion Force is strongest in those molecules. They shall have the highest boiling points.

Examples: (SynQuest)

  • C₂H₆ (ethane)- b.p. -88 °C
  • C₄H₁₀ (butane)- b.p. -1 °C ~ 1 °C
  • C₆H₁₄ (hexane)- b.p. 68 °C ~ 70 °C
<h2>(b) </h2>

The conclusion in (a) likely holds molecules in the same homologous series.

<h3>Explanation</h3>

Molecules in the same homologous series have the same types and numbers of functional groups. However, they differ only in the number of repeating units (-CH₂ in this case) that they contains. The number of such units in each molecule is directly related to the length of its carbon backbone.

Functional groups introduce extra types of forces between the molecules. For instance:

  • Halogens, e.g., Cl, forms polar bonds with carbon. In most cases they make the molecule polar enough to form dipole-dipole interactions.
  • Hydroxyl groups -OH can lead to hydrogen bonds between the molecules.

Functional groups tend to have similar effects on the b.p. in the same homologous series. The extra interaction due to the functional groups stays generally the same. Trends in the strength of dispersion forces likely follow the reasoning in (a). There shall be a similar conclusion. Molecules with the longest backbone in the same homologous series would have the highest b.p.

Example: (FooDB)

  • CH₃COOH (ethanoic acid)- b.p. 118 °C
  • C₂H₅COOH (butanoic acid)- b.p. 163.5 °C
  • C₃H₇COOH (hexanoic acid)- b.p. 205 °C

Note, that it's only the number of repeating units in the carbon backbone that differs. Functional groups shall be on the similar positions on members of the series. For instance, 1,1-dichloroethane Cl₂-CH-CH₃ and 1,1-dichloropropane Cl₂-CH-CH₂-CH₃ are on the same series; whereas 1,3-dichloropropane Cl-CH₂-CH₂-CH₂-Cl is not.

7 0
3 years ago
Can someone solve this problem 5
Westkost [7]

Answer:

2

Step-by-step explanation:

A. Moles before mixing

<em>Beaker I: </em>

Moles of H⁺ = 0.100 L × 0.03 mol/1 L

                   = 3 × 10⁻³ mol

<em>Beaker II: </em>

Beaker II is basic, because [H⁺] < 10⁻⁷ mol·L⁻¹.

        H⁺][OH⁻] = 1 × 10⁻¹⁴   Divide each side by [H⁺]

             [OH⁻] = (1 × 10⁻¹⁴)/[H⁺]

             [OH⁻] = (1 × 10⁻¹⁴)/(1 × 10⁻¹²)

             [OH⁻] = 0.01 mol·L⁻¹

Moles of OH⁻ = 0.100 L × 0.01 mol/1 L

                      = 1 × 10⁻³ mol

B. Moles after mixing

                 H⁺    +    OH⁻   ⟶ H₂O

I/mol:      3 × 10⁻³   1 × 10⁻³

C/mol:   -1 × 10⁻³  -1 × 10⁻³

E/mol:    2 × 10⁻³          0

You have more moles of acid than base, so the base will be completely neutralized when you mix the solutions.

You will end up with 2 × 10⁻³ mol of H⁺ in 200 mL of solution.


C. pH

 [H⁺] = (2 × 10⁻³ mol)/(0.200 L)

        = 1 × 10⁻² mol·L⁻¹

 pH = -log[H⁺ ]

       = -log(1 × 10⁻²)

       = 2

6 0
3 years ago
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