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larisa86 [58]
4 years ago
15

The moons appearance from the earth changes throughout its orbital path because the amount of .... varies ?

Chemistry
1 answer:
Mkey [24]4 years ago
3 0
The amount of reflected sunlight varies. There really isn't any such thing as moonlight. The light we perceive to be produced by the moon is really just sunlight! As the moon moves around the planet, we can see differing amounts of reflected light that we categorize as phases.

<span>I hope this helps! Let me know if you have anymore questions, and have a great day! :)</span>
You might be interested in
Select all of the statements that are true about a buffer solution. Select one or more: a. A buffer solution reacts with basic s
zhuklara [117]

Answer:

a. A buffer solution reacts with basic solutions.

c. A buffer solution reacts with acidic solutions.

e. A buffer solution resists small changes in pH

Explanation:

1. Buffer questions

a, c, and e are TRUE. A buffer resists a change in the pH when small amounts of a strong acid or base are added to it.

b is wrong. A buffer can have a pH of 7, but it can also have many other pH values.

d is wrong. Most buffers are colourless, and they resist a change in pH.

2. Titration curves

The solution is the best buffer at the mid-point of the titration curve.

In the figure below, the equivalence point is at 13 mL, so the mid-point is at 6.5 mL.  

The solution is buffered at pH 3.2.

However, the solution is a buffer at any point in the range pH = 3.2 ± 1.

That would be in the range of 1 mL to 12 mL.

The buffering ability becomes worse the further you are from the mid-point of the titration.

5 0
4 years ago
Which phrase correctly describes the particles that make up water vapor?
Alexxx [7]

Answer:

i believe its A but im not sure

its the only one that makes any sense

plz mark  brainliest

\

Explanation:

7 0
3 years ago
The vapor pressure of benzene at 298 K is 94.4 mm of Hg. The standard molar Gibbs free energy of formation of liquid benzene at
horsena [70]

Answer:

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

Explanation:

Bringing out the parameters mentioned in the question;

Vapor pressure = 94.4 mm of Hg

The vaporization reaction is given as;

C₆H₆(l) ⇄ C₆H₆(g)

Equilibrium in terms of activities is given by:

K = a(C₆H₆(g)) / a(C₆H₆(l))

Activity of pure substances is one:

a(C₆H₆(l)) = 1

Assuming ideal gas phase activity equals partial pressure divided by total pressure. At standard conditions

K = p(C₆H₆(g)) / p°

Where p° = 1atm = 760mmHg standard pressure

We now have;

K = 94mmHg / 760mmHg = 0.12421

Gibbs free energy is given as;

ΔG = - R·T·ln(K)

where R = gas constant = 8.314472J/molK

So ΔG° of vaporization of benzene is:

ΔvG° = - 8.314472 · 298.15 · ln(0.12421)

ΔvG° = 5171J/mol = 5.2kJ/mol  

Gibbs free energy change of reaction = Gibbs free energy of formation of products - Gibbs free energy of formation of reactants:

ΔvG° = ΔfG°(C₆H₆(g)) - ΔfG°(C₆H₆(l))

Hence:

ΔfG°(C₆H₆(g)) = ΔvG°+ ΔfG°(C₆H₆(l))

ΔfG°(C₆H₆(g)) = 5.2kJ/mol + 124.5kJ/mol

ΔfG°(C₆H₆(g)) = 129.7kJ/mol

6 0
4 years ago
The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

5 0
3 years ago
100 points
Mariana [72]

Answer:

Where is the puzzle!??

I surely can help you

5 0
3 years ago
Read 2 more answers
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