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Sholpan [36]
3 years ago
7

How many kilojoules of heat are released when 0.72 mole of oxygen gas are used to combust methane?

Chemistry
1 answer:
KonstantinChe [14]3 years ago
6 0

0.72 mole of oxygen would produce 320.4 kJ of heat.

<u>Explanation:</u>

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (ℓ) + 890kJ

According to the equation,

2 moles of O₂ produces 890 kJ of heat

So, 0.72 moles of O₂ will produce:

= \frac{890}{2} X 0.72\\\\= 445 X 0.72\\\\= 320.4 kJ

Therefore, 0.72 mole of oxygen would produce 320.4 kJ of heat.

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The answer is atom because it cannot be broken down into small substances
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Ideal gas law, Please help me it's due soon
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is the equation of state of a hypothetical ideal gas

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A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol
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Answer:

0.44 moles

Explanation:

Given that :

A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.

The equilibrium constant K_c=  \dfrac{[CO][H_2]}{[H_2O]}

The equilibrium constant  K_c=  \dfrac{(0.17 )(0.17)}{0.74}

The equilibrium constant K_c=  0.03905

Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.

The equation for the reaction is :

H_2 + \dfrac{1}{2}O_2 \to H_2O \\ \\ 0.17 \ \ \ \ \  \ \ \ \ \to0.17

Total mole of water now = 0.74+0.17

Total mole of water now = 0.91 moles

Again:

K_c=  \dfrac{[CO][H_2]}{[H_2O]}

0.03905 =  \dfrac{[0.17+x][x]}{[0.91 -x]}

0.03905(0.91 -x) = (0.17 +x)(x)

0.0355355 - 0.03905x = 0.17x + x²

0.0355355 +0.13095 x -x²

x² - 0.13095 x - 0.0355355 = 0

By using quadratic formula

x = 0.265  or   x = -0.134

Going by the value with the positive integer; x = 0.265 moles

Total moles of CO in the flask when the system returns to equilibrium is :

= 0.17 + x

= 0.17 + 0.265

= 0.435 moles

=0.44 moles (to two significant figures)

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3 years ago
What is the purpose of a
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2 years ago
If 13.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
Paul [167]

6.349 g mass of anhydrous magnesium sulfate will remain.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Molar mass MgSO₄.7 H₂O = 246.52 g/mol

Moles =\frac{mass}{molar \;mass}

Moles =\frac{13.0 g}{246.52}

0.0527 moles

Molar mass MgSO₄ = 120.4 g/mol

Mass of anhydrous magnesium sulfate :

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Learn more about moles here:

brainly.com/question/8455949

#SPJ1

4 0
1 year ago
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