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Yuliya22 [10]
3 years ago
15

Help! ASAP! Please on the questions

Chemistry
2 answers:
KiRa [710]3 years ago
8 0

Answer:

62.15 m (squared)

Explanation:

We know that the Area of the square is 5.5 x 4= 22

so 90 degree to 30 degree is divide by 3

so u divide 22 by 3 and get 7.3 wich is the semi circle thing

now 7.3 x 5.5 = 40.15

22 + 40.15 = 62.15 m (squared)

Look im not really sure but I think this is the answer

<em><u>Please mark as brainliest</u></em>

Have a great day, be safe and healthy

Thank u  

XD

monitta3 years ago
8 0

Answer:

1. The area of the rectangle is 22 but I'm not sure the area of the shape above sorry.

A. 28 m

B. 672 m

Explanation:

A. 6 + 6 + 8 + 8 = 28 or 2(6+8) = (2)14 = 28

B. 28 * 24 = 672

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How does an increase in temperature affect the equilibrium concentration of the underlined substance and K for each of the follo
Darya [45]

Explanation:

Due to the positive value of the change in temperature, this is an endothermic reaction.

Since the forward reaction is endothermic, increasing the temperature increases the equilibrium constant (k).

In an equilibrium system, the position of the equilibrium will move in a way to annul the change made to the system. An increase in temperature for an endothermic reaction would favour the reaction, leading to increase in amount of products and decrease in amount of reactants.

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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

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