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3 years ago

Which diagram shows a wave with the highest frequency? Chemistry question.

Chemistry

1 answer:

notsponge [240]3 years ago

8 0

The correct answer is D

Explanation:

Wave frequency is mainly determined by the number of waves that pass through a specific point. In a diagram, this can be found by analyzing the number of crests (top of the wave) and the space between them. For example, wave B is the one with the lowest frequency because there is only one crest and this shows only one wave passing at a specific point. On the opposite, wave D is the one with the highest frequency because this shows multiple crests and this indicates the frequency is high or that many waves pass through a specific point in a short time.

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Determine the mole fraction of the solute if you have 3.4 grams KCl in 8.1 grams H2O

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3 years ago

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3 years ago

Can someone solve this problem 5

Westkost [7]

Answer:

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

In a aqueous solution of 4-chlorobutanoic acid , what is the percentage of -chlorobutanoic acid that is dissociated

lord [1]

Answer:

Explanation:

Let assume that the missing aqueous solution of 4-chlorobutanoic acid = 0.76 M

Then, the dissociation of 4-chlorobutanoic acid can be expressed as:

$\mathsf{C_3H_6ClCO_2H }$ ⇄ $\mathsf{C_3H_6ClCO_2^-}$ + $\mathsf{H^+}$

The ICE table can be computed as:

$\mathsf{C_3H_6ClCO_2H }$ ⇄ $\mathsf{C_3H_6ClCO_2^-}$ + $\mathsf{H^+}$

Initial 0.76 - -

Change -x +x +x

Equilibrium 0.76 - x x x

$K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}$

$K_a = \dfrac{[x] [x]}{ [0.76-x]}$

where:

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}$

however, the value of x is so negligible:

0.76 -x = 0.76

Then:

$3.02*10^{-5}*0.76 = x^2$

$x=\sqrt{3.02*10^{-5}*0.76 }$

x = 0.00479 M

∴

$x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=}$ 0.00479 M

$\mathsf{C_3H_6ClCO_2H }$ = (0.76 - 0.00479) M

= 0.75521 M

Finally, the percentage of the acid dissociated is;

= ( 0.00479 / 0.76) × 100

= 0.630 M

7 0

3 years ago

Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 8.97×10−3. Express your answer to thr

puteri [66]

Answer: The standard cell potential (E∘) for the reaction $X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$ is -0.121 V

Explanation:

The reaction is:

$X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = Standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = 298 K

K= equilibrium constant = $8.97\times 10^{-3}$

Putting values in above equation, we get:

$\Delta G^0=-(8.314J/Kmol\times 298K\times \ln (8.97\times 10^{-3})\\\\\Delta G^0=11678.9J/mol$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 1

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = ?

Putting values in above equation, we get:

$11678.9J/mol=-1\times 96500\times E^0_{cell}$

$\frac{11678.9J/mol}{-96500}=E^0_{cell}$

$-0.121V=E^0_{cell}$

Thus standard cell potential (E∘) for the reaction $X(s)+Y^+(aq)\rightarrow X^+(aq)+Y(s)$ is -0.121 V

5 0

4 years ago