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3 years ago

Which evidence supports the claim that matter can bend light waves?

Chemistry

2 answers:

solmaris [256]3 years ago

8 0

Answer:d

Explanation:I’m not 100% sure how to explain it but I’m almost for sure it’s d.

pantera1 [17]3 years ago

8 0

Answer:

Explanation:

I think it would be D. Refraction is an example of matter (being water) bending light as it travels through another medium (from air)

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What will an object moving at a constant velocity do?

Len [333]

Answer:B

Explanation: an object in motion will stay in motion until acted on by another force and an object in rest will stay in test until acted on by another force

3 0

3 years ago

In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined

faltersainse [42]

Answer:

For A: The standard cell potential of the reaction is 4.4 V

For B: The standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C: The reaction is spontaneous as written.

Explanation:

For A:

The given chemical reaction follows:

$2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)$

The given half reaction follows:

Oxidation half reaction: $Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V$ ( × 2)

Reduction half reaction: $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=1.36-(-3.04)=4.4V$

Hence, the standard cell potential of the reaction is 4.4 V

For B:

Relationship between standard Gibbs free energy and standard electrode potential follows:

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

where,

n = number of electrons transferred = $2mol\text{ e}^-$

F = Faradays constant = $96500J/V.mol\text{ e}^-$

$E^o_{cell}$ = standard cell potential = 4.4 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J$

Hence, the standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C:

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

7 0

4 years ago

Chloroacetic acid, ClCH2COOH, has a pKa of 2.87. What are [H3O+], pH, [ClCH2COO−], and [ClCH2COOH] in 1.55 M ClCH2COOH?

Mademuasel [1]

Answer:

See explanation below

Explanation:

first to all, this is an acid base reaction where the chloroacetic acid is being dissociated in water. Therefore, is an equilibrium reaction.

ClCH₂COOH has a pKa of 2.87 so, the Ka would be:

pKa = -logKa ---> Ka = antlog(-pka)

Ka = antlog(-2.87)

Ka = 1.35x10⁻³

Now that we know the Ka, we need to write the chemical reaction and then, an ICE chart:

ClCH₂COOH + H₂O <----------> ClCH₂COO⁻ + H₃O⁺ Ka = 1.35x10⁻³

i) 1.55 0 0

c) -x +x +x

e) 1.55-x x x

Writting now the equilibrium reaction:

Ka = [H₃O⁺] [ClCH₂COO⁻] / [ClCH₂COOH]

Replacing the values of the chart:

1.35x10⁻³ = x² / 1.55-x

1.35x10⁻³(1.55-x) = x²

2.0925x10⁻³ - 1.35x10⁻³x = x²

x² + 1.35x10⁻³x - 2.0925x10⁻³ = 0 --> a = 1; b = 1.35x10⁻³x; c = 2.0925x10⁻³

From here we use the general equation for solve x in a quadratic equation which is:

x = -b±√(b² - 4ac) / 2a

Replacing the values we have:

x = -1.35x10⁻³ ±√(1.35x10⁻³)² - 4*1*(-2.0925x10⁻³) / 2

x = -1.35x10⁻³ ±√(8.37x10⁻³) / 2

x = -1.35x10⁻³ ± 0.091 / 2

x1 = -1.35x10⁻³ + 0.091 / 2 = 0.045 M

x2 = -1.35x10⁻³ - 0.091 / 2 = -0.046 M

In this case, we will take the positive value of x, in this case, x1.

With this value, the equilibrium concentrations are the following:

[H₃O⁺] = [ClCH₂COO⁻] = 0.045 M

[ClCH₂COOH] = 1.55 - 0.045 = 1.505 M

Finally the pH:

pH = -log[H₃O⁺]

pH = -log(0.045)

pH = 1.35

4 0

3 years ago

The air bags in automobiles were once inflated by nitrogen gas generated by the rapid decomposition of

Basile [38]

Answer:

1.57 mol NaN₃

Explanation:

2 NaN₃ (s) → 2 Na (s) + 3 N₂ (g)

First we use PV=nRT to calculate the number of N₂ moles that need to be produced:

P = 1.07 atm

V = 53.4 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 23.7 °C ⇒ 23.7 + 273.16 = 296.86 K

Inputing the data:

1.07 atm * 53.4 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.86

And solving for n:

n = 2.35 mol N₂

Finally we convert N₂ moles into NaN₃ moles, using the stoichiometric coefficients of the balanced reaction:

2.35 mol N₂ * $\frac{2molNaN_3}{3molN_2}$ = 1.57 mol NaN₃

4 0

3 years ago

The boiling point of a substance in City A is found to be 145°C the boiling point of th same substance in Xity B is 141°C which

DochEvi [55]

City B. Higher altitudes have lower boiling points due to lower atmospheric pressure at higher altitudes

4 0

3 years ago