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djverab [1.8K]
3 years ago
7

The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kJ. You may want to re

ference (Pages 381 - 385) Section 9.6 while completing this problem.
Chemistry
1 answer:
aivan3 [116]3 years ago
6 0

This is an incomplete question, here is a complete question.

The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kJ. You may want to reference (Pages 381 - 385) Section 9.6 while completing this problem. If the change in enthalpy is -5074.2 kJ, how much work is done during the combustion? Express the work in kilojoules to three significant figures.

Answer : The work done during the combustion is, 9.9 kJ

Explanation :

Formula used :

\Delta H=\Delta E+w

where,

w = work done = ?

\Delta H = change in enthalpy  = -5074.2 kJ

\Delta E = change in internal energy  = -5084.1 kJ

R = gas constant = 8.314 J/mol.K

Now put all the given values in the above formula, we get:

-5074.2kJ=-5084.1kJ+w

w=9.9kJ

Thus, the work done during the combustion is, 9.9 kJ

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Answer:

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Explanation:

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Lt takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. what is the half-life of radium-230?
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Radioactive decay => C = Co { e ^ (- kt) |

Data:

Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min

Time conversion: 4 hr 39 min = 4.65 hr

1) Replace the data in the equation to find k

C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}

=> k = ln {Co / C} / t =  ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719

2) Use C / Co  = 1/2 to find the hallf-life

C / Co = e ^ (-kt) => -kt = ln (C / Co)

=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k

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4 years ago
URGENT!!! What is the balanced form of this equation?
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To balance out the H^+ on the reactant side, we write H_2O on the product side.

Balancing out the following reaction gives us:

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If 100.0g of nitrogen is reacted with 100.0g of hydrogen, what is the excess reactant? What is the limiting reactant? Show your
Drupady [299]

N₂ : limiting reactant

H₂ : excess reactant

<h3>Further explanation</h3>

Given

mass of N₂ = 100 g

mass of H₂ = 100 g

Required

Limiting reactant

Excess reactant

Solution

Reaction

<em>N₂+3H₂⇒2NH₃</em>

mol N₂(MW=28 g/mol) :

\tt mol=\dfrac{mass}{MW}=\dfrac{100}{28}=3.571

mol H₂(MW= 2 g/mol) :

\tt mol=\dfrac{100}{2}=50

A method that can be used to find limiting reactants is to divide the number of moles of known substances by their respective coefficients, and small or exhausted reactans become a limiting reactants

From the equation, mol ratio N₂ : H₂ = 1 : 3, so :

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