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Greeley [361]
3 years ago
10

How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr?

Physics
1 answer:
alexdok [17]3 years ago
3 0

Heat energy released is - 31.4 kJ.

<u>Explanation:</u>

CO₍g₎ + 2H₂₍g₎ → CH₃OH₍l₎

Volume of CO, V ₍CO₎ = 15 L = 0.015 m³

Pressure = 112 kPa = 112,000 Pa

T= 85 ⁰C = 85 + 273 = 358 K

As per ideal gas law,

PV = nRT

n = \frac{PV}{RT}

   = 112000 × 0.015 / 8.314 × 358

 n(CO) = 0.56 moles

Volume of H₂ = 14.4 L = 0.0144 m³

Pressure = 744 torr = 99191.84 Pa

T= 75⁰C + 273 K = 348 K

n(H₂) = 99191.84 ×0.0144 m³ / 8.314 ×348 K

     = 0.49 moles of H₂

The above calculation shows that hydrogen is the limiting reagent.

n(CH₃OH) = n(H₂) /2

        = 0.49/2 = 0.245 mol

ΔH₍rxn₎ = ΔH₍f₎ (CH₃OH) - ΔH₍f₎(CO)

           = -238.6 -(-110.5)

           = -128.1 kJ/mol

Now we have to multiply ΔH₍rxn₎ with the moles of methanol.

E = ΔH₍rxn₎ × n(CH₃OH) = -128.1 ×0.245

        = - 31.4 kJ.

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Please help Math Phys
Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land.  Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

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