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3 years ago

15

What is the standard enthalphy change ΔHo, for the reaction represented above? (ΔHof of C2H2(g) is 230 kJ mol-1; (ΔHof of C6H6(g

) is 83 kJ mol-1;)

1 answer:

wolverine [178]3 years ago

6 0

Answer:

-608KJ/mol

Explanation:

3 C2H2(g) -> C6H6(g)

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= ΔHC6H6 - 3ΔHC2H2

ΔHrxn = 83 - 3(230)

ΔHrxn = -608

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otez555 [7]

Answer:

P2≈393.609Kpa so I think the answer is 394 kPa

Explanation:

PV=mRT Ideal Gas Law

m and R are constant because they dont change for the problem. That means

PV/T=mR = constant

so P1*V1/T1=P2*V2/T2 and note that the temperatures are in absolute temperatures (Kelvin) because you can't divide by zero.

So P2 = P1*V1*T2/(V2*T1) = 101325 Pa * 700 mL * 303K/(200 mL*273K)

P2 = 393609 Pa

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Elena L [17]

Answer:

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Explanation:

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4 years ago

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The element bromine is a reddish-brown liquid at room temperature, with a density of 3.103g/mL. What is the mass, in grams, of 1

Anastaziya [24]

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3 years ago

What types of fields does moving electricity produce?

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3 years ago

How many moles of NH3 can be produced from 12.0 mol of H2 and excess N2? Express your answer numerically in moles. View Availabl

VladimirAG [237]

Answer:

A) 8.00 mol NH₃

B) 137 g NH₃

C) 2.30 g H₂

D) 1.53 x 10²⁰ molecules NH₃

Explanation:

Let us consider the balanced equation:

N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)

Part A

3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:

$12.0molH_{2}.\frac{2molNH_{3}}{3molH_{2}} =8.00molNH_{3}$

Part B:

1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:

$4.04molN_{2}.\frac{2molNH_{3}}{1molN_{2}} .\frac{17.0gNH_{3}}{1molNH_{3}} =137gNH_{3}$

Part C:

According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:

$13.02gNH_{3}.\frac{6.00gH_{2}}{34.0gNH_{3}} =2.30gH_{2}$

Part D:

6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂:

$7.62 \times 10^{-4} gH_{2}.\frac{2molNH_{3}}{6.00gH_{2}} .\frac{6.02\times 10^{23}moleculesNH_{3} }{1molNH_{3}}=1.53\times10^{20}moleculesNH_{3}$

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4 years ago