6500 in 2 significant figures

11 A student combined 1.2 g of H2 gas and 5.4 g of He gas in a 2.5 L closed container. What was the pressure exerted by the H2 g

blagie [28]

The correct answer is 5.8 atm.

7 0

3 years ago

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For the following reaction, 4.64 grams of oxygen gas are mixed with excess benzene (C6H6). The reaction yields 3.95 grams of car

Reil [10]

Answer:

Theoretical yield for CO₂ is 5.10g

Explanation:

Reaction: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(g)

We convert the mass of oxygen to moles:

4.64 g /32 g/mol = 0.145 moles of O₂

Let's find out the 100% yield reaction of CO₂ (theoretical yield)

Ratio is 15:12. So let's make this rule of three:

15 moles of O₂ can produce 12 moles of CO₂

Therefore 0.145 moles of oxygen will produce (0.145 . 12) /15 = 0.116 moles

We convert the moles to mass: 0.116 mol . 44 g / 1mol = 5.10 g

6 0

3 years ago

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Which of these energy resources are renewable? Select all that apply.

xz_007 [3.2K]

The correct options are option B and D, that is, wind power and geothermal energy.

The form of energy which is gathered from renewable resources is known as renewable energy. This form of energy gets restocked on a human timescale like wind, sunlight, tides, waves, rain, and geothermal heat.

Renewable energy usually provides energy in four essential fields, that is, air and water heating/cooling, electricity generation, rural energy services, and transportation.

4 0

3 years ago

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A sample consisting of 1.00 mol of perfect gas molecules at 27 °C is expanded isothermally from an initial pressure of 3.00 atm

Evgesh-ka [11]

Answer:

a) reversibly

ΔU = 0

q = 2740.16 J

w = -2740.16 J

ΔH = 0

ΔS(total) = 0

ΔS(sys) =9.13 J/K

ΔS(surr) = -9.13 J/K

b) against a constant external pressure of 1.00 atm

ΔU = 0

w = -1.66 kJ

q = 1.66 kJ

ΔH = 0

ΔS(sys) = 9.13 J/K

ΔS(surr) = -5.543 J/K

ΔS(total) = 3.587 J/K

Explanation:

Step 1: Data given:

Number of moles = 1.00 mol

Temperature = 27.00 °C = 300 Kelvin

Initial pressure = 3.00 atm

Final pressure = 1.00 atm

The gas constant = 8.31 J/mol*K

(a) reversibly

Step 2: Calculate work done

For ideal gases ΔU depends only on temperature. So as it is an isothermal (T constant).

Since the temperature remains constant:

ΔU = 0

ΔU = q + w

q = -w

w = -nRT ln (Pi/Pf)

⇒ with n = the number of moles of perfect gas = 1.00 mol

⇒ with R = the gas constant = 8.314 J/mol*K

⇒ with T = the temperature = 300 Kelvin

⇒ with Pi = the initial pressure = 3.00 atm

⇒ with Pf = the final pressure = 1.00 atm

w =- 1*8.314 *300 * ln(3)

w = -2740.16 J

q = -w

q = 2740.16 J

Step 3: Calculate change in enthalpy

Since there is no change in energy, ΔH = 0

Step 4: Calculate ΔS

for an isothermal process

ΔS (total) = ΔS(sys) + ΔS(surr)

ΔS(sys) = -ΔS(surr)

ΔS(sys) = n*R*ln(pi/pf)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -9.13 J/K

ΔS (total) = ΔS(sys) + ΔS(surr) = 0

(b) against a constant external pressure of 1.00 atm

Step 1: Calculate the work done

w = -Pext*ΔV

w = -Pext*(Vf - Vi)

⇒ with Vf = the final volume

⇒ with Vi = the initial volume

We have to calculate the final and initial volume. We do this via the ideal gas law P*V=n*R*T

V = (n*R*T)/P

Initial volume = (n*R*T)/Pi

⇒ Vi = (1*0.08206 *300)/3

⇒ Vi = 8.206 L

Final volume = (n*R*T)/Pf

⇒ Vf = (1*0.08206 *300)/1

⇒ Vf = 24.618 L

The work done w = -Pext*(Vf - Vi)

w = -1.00* ( 24.618 - 8.206)

w = -16.412 atm*L

w = -16 .412 *(101325/1atm*L) *(1kJ/1000J)

w = -1662.9 J = -1.66 kJ

Step 2: Calculate the change in internal energy

ΔU = 0

q = -w

q = 1.66 kJ

ΔH = 0 because there is no change in energy

Step 3: Calculate ΔS

ΔS(sys) = n*R*ln(3)

ΔS(sys) = 1.00 * 8.314 * ln(3)

ΔS(sys) = 9.13 J/K

ΔS(surr) = -q/T

ΔS(surr) = -1662.9J/300K

ΔS(surr) = -5.543 J/K

ΔS(total) = ΔS(surr) +ΔS(sys) = -5.543 J/K + 9.13 J/K = 3.587 J/K

4 0

4 years ago

Consider the following reaction where Kc = 34.5 at 1150 K:2SO2(g) + O2(g) 2SO3(g)A reaction mixture was found to contain 4.77×10

sukhopar [10]

Answer:

Explanation:

[ so₃] = 4.37 x 10⁻²

[so₂] = 4.77 x 10⁻²

[ o₂] = 4.55 x 10⁻²

Qc = (4.37)²x10⁻⁴ /(4.77)².(4.55) x 10⁻⁶ =18.44

Qc is less than Kc hence in order to reach equilibrium more of so₃ will be produced . Statement 1 is true.

Kc is always constant . Statement 2 is false.

Statement 3 is false because statement 1 is true.

Qc Is smaller than Kc . So statement 4 is false.

The reaction is not in equilibrium. Statement 5 is false.

3 0

4 years ago