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3 years ago

9

Please Help me, I only have today to do this. I don't want to fail

1 answer:

leva [86]3 years ago

3 0

Answer:

Part 2: for waxing crescent first quarter and waxing gibbous

for waning crescent new moon and waxing crescent moon

ForPart 1: As the moon revolves around Earth each month, it reflects the light of the sun. August 10: Full moon August 18: Third quarter August 23: Either waxing crescent or waning crescent idk. August 23 I'm not sure which crescent so choose one of them also use spinbot so you don't copy word from word and get in trouble hope this helps ;D

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Reactioneaza total 22 kg calcar de puritate 80 % cu o solutie de acid clorhidric de concentratie 30 %. Stiind ca rezulta clorura

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Answer:

De la Alex;)(da alex ăla de e din cls cu tn:)

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3 years ago

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padilas [110]

The best way to do this is to google search each question, or look it up on quizlet.com

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Aleksandr [31]

Answer:

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Explanation:

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3 years ago

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: I2(s) and Br-(aq) Give your answer usi

insens350 [35]

Explanation:

Formula to calculate standard electrode potential is as follows.

$E^{o}_{cell} = E^{0}_{cathode} - E^{0}_{anode}$

= 0.535 - 1.065

= - 0.53 V

Also, it is known that relation between $E^{o}_{cell}$ and K is as follows.

$E^{o}_{cell} = \frac{RT}{nF} \times ln K$

ln K = $\frac{nFE^{0}_{cell}}{RT}$

Substituting the given values into the above formula as follows.

ln K = $\frac{nFE^{0}_{cell}}{RT}$

= $\frac{2 \times 96485 C mol^{-1} \times -0.53 V}{8.314 l atm/mol K \times 298 K} \times \frac{1 J}{1 V C}$

ln K = -41.28

K = $e^{-41.28}$

= $1 \times 10^{-18}$

Thus, we can conclude that the value of the equilibrium constant for the given reaction is $1 \times 10^{-18}$ .

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3 years ago

4Al(s)+3O2(g)__> 2Al2O3(s)

nasty-shy [4]

What do you want us to work out?

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3 years ago