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3 years ago

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SAT

2 answers:

satela [25.4K]3 years ago

7 0

WHDBDHWIWHH;&2938;)3?2?

allsm [11]3 years ago

5 0

Alalallaaoaoapapappapapapapapapappapapapapapapapapapa

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At what temperature teq do the forward and reverse corrosion reactions occur in equilibrium?

Vedmedyk [2.9K]

Standard Gibb's free energy at 298K and 5975K is 3.16×10⁶J and -3.83×10⁵J respectively, and temperature at equilibrium state is 5362.3 K.

<h3>What is the standard Gibb's free energy?</h3>

The complete question is in the brainly link at the end. We can say that;

Standard Gibb's free energy of any reaction is gotten from the formula;

∆G° = ∆H° - T∆S°

where;

∆G° is change in free energy

∆H° is change in enthalpy = 3352 kJ = 3352 × 10³J

∆S° is change in entropy = 625.1 J/K

T is absolute temperature = 298 K

Plugging in the relevant values into the gibbs energy equation gives;

∆G° = (3352 × 10³) - (298 × 625.1)

∆G° = 3.16 × 10⁶J

Gibb's free energy at temperature of 5,975 K is;

∆G° = (3,352 × 10³) - (5,975 × 625.1)

∆G° = -3.83 × 10⁵J

At the equilibrium state, the value of Gibb's free energy is zero. Thus we can calculate the temperature as;

0 = (3352 × 10³)J - (T × 625.1 J/K)

Solving for T gives;

T = 5362.3 K

Read more about Standard gibbs free energy at; brainly.com/question/17205875

5 0

3 years ago

Greta says that 50% of a number will always be less than 75% of any other number.

Galina-37 [17]

Answer:that is tru

Explanation:

3 0

3 years ago

1/4V= 2 + 3

Mandarinka [93]

Answer:

V=5:(-1/4)

V=20

6*V-17+18-19+20

6*20-17+18-19+20=122

4 0

3 years ago

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I wish my cousin was here with me.

Travka [436]

Answer:

aw... I wish mine were too. I never get to see any of them :(

Explanation:

I hope you get to see them soon!

3 0

2 years ago

Identify the reference angle es001-1. Jpg for each given angle, es001-2. Jpg. Es001-3. Jpg degrees. Es002-1. Jpg degrees. Es003-

julia-pushkina [17]

The reference angle is the smallest angle between the terminal side and the x-axis.

$(1)\ \theta = 300^o$

300 degrees is in the 4th quadrant.

So, the reference angle is

$\alpha = 360 - \theta$

$\alpha = 360 - 300$

$\alpha = 60^o$

Hence, the reference angle of 300 degrees is 60 degrees

$(2)\ \theta = 225^o$

225 degrees is in the 3rd quadrant.

So, the reference angle is

$\alpha = \theta - 180$

$\alpha = 225 - 180$

$\alpha = 45^o$

Hence, the reference angle of 225 degrees is 45 degrees

$(3)\ \theta = 480^o$

480 degrees is in the 2nd quadrant (i.e. 480 - 360 = 120)

So, the reference angle is

$\alpha = 180 - \theta$

$\alpha = 180 - 120$

$\alpha = 60^o$

Hence, the reference angle of 480 degrees is 60 degrees

$(2)\ \theta = -210^o$

-210 degrees is in the 2nd quadrant (i.e. 360 - 210 = 150)

So, the reference angle is

$\alpha = 180 - \theta$

$\alpha = 180 - 150$

$\alpha = 30^o$

Hence, the reference angle of -210 degrees is 30 degrees

Read more about reference angles at:

brainly.com/question/9601871

7 0

2 years ago

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