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trasher [3.6K]
3 years ago
10

If the amplitude of ocean waves increases by a factor of 1.1, by how much does the energy increase?

Chemistry
1 answer:
klasskru [66]3 years ago
6 0

Answer:

The energy increases by a factor of 1.2.

Explanation:

The energy (<em>E</em>) of an ocean wave is directly proportional to the square of its amplitude (<em>A</em>).

<em>E</em> ∝ <em>A</em>² or <em>E = kA</em>²

If you have two waves with amplitudes <em>A</em>₁ and <em>A</em>₂, then

\frac{E_{2} }{E_{1} } = \frac{A_{2}^{2} }{A_{1}^{2} } = (\frac{A_{2} }{A_{1} }})^{2}

If <em>A</em>₂ = 1.1<em>A</em>₁, then

E_{2} = E_{1} \times (\frac{1.1A_{1} }{A_{1} })^{2} = (1.1)^{2} = 1.2

The energy increases by a factor of 1.2.

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15.142338 mol/kg

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(1 mole) / (66.04 grams)

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Determine whether each description represents a genotype or a phenotype.
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A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so
Xelga [282]

Answer : The equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

Explanation :

First we have to calculate the initial moles of Fe^{3+} and SCN^-.

\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}

\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol

and,

\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}

\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol

The given balanced chemical reaction is,

Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)

Since 1 mole of Fe^{3+} reacts with 1 mole of SCN^- to give 1 mole of FeSCN^{2+}

The limiting reagent is, SCN^-

So, the number of moles of FeSCN^{2+} = 0.0020 mmole

Now we have to calculate the concentration of FeSCN^{2+}.

\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M

Using Beer-Lambert's law :

A=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution

l = path length

\epsilon = molar absorptivity coefficient

\epsilon and l are same for stock solution and dilute solution. So,

\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}

For trial solution:

The equilibrium concentration of SCN^- is,

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{initial} = 0.00050 M

Now calculate the [FeSCN^{2+}].

C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M

Now calculate the concentration of SCN^-.

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)

[SCN^-]_{eqm}=4.58\times 10^{-8}M

Therefore, the equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

5 0
3 years ago
Acetylene is hyrodgenated to form ethane. The feed to the reactor contains 1.60 mol H2/mol C2H2. The reaction proceeds to comple
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Answer:

Explanation:

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b )

yield ratio = mole C₂H₆ formed / mole H₂ reacted ) = 0.8 / 1.6 = 1/2 = 0.5 .

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