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tangare [24]
4 years ago
15

20cm³ of methane gas was burnt in 20cm³ of oxygen. write a balanced chemical equation for the reaction and determine which of th

e gases were in excess​
Chemistry
1 answer:
solniwko [45]4 years ago
7 0

Answer:

CH4 +O2 => CO2 + 2 H2O

None were in excess

Explanation:

Equation for the reaction is,

CH4 + O2 =>CO2 +2 H2O

No of moles of CH4 = (20 /1000)/24 =0.02 /24 = 0.00083

No of moles of O2 =20 /24000 = 0.0083

CH4 : O2 = 1:1

THEREFORE

None of the gases were in excess.

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Answer:

selenium dioxide

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Given the following data:
bagirrra123 [75]

176.0 \; \text{kJ} \cdot \text{mol}^{-1}

As long as the equation in question can be expressed as the sum of the three equations with known enthalpy change, its \Delta H can be determined with the Hess's Law. The key is to find the appropriate coefficient for each of the given equations.

Let the three equations with \Delta H given be denoted as (1), (2), (3), and the last equation (4). Let a, b, and c be letters such that a \times (1) + b \times (2) + c \times (3) = (4). This relationship shall hold for all chemicals involved.

There are three unknowns; it would thus take at least three equations to find their values. Species present on both sides of the equation would cancel out. Thus, let coefficients on the reactant side be positive and those on the product side be negative, such that duplicates would cancel out arithmetically. For instance, 3 + (-1) = 2 shall resemble the number of \text{H}_2 left on the product side when the second equation is directly added to the third. Similarly

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Thus

a = -1/2\\b = 1/2\\c = -1/2 and

-\frac{1}{2} \times (1) + \frac{1}{2} \times (2) - \frac{1}{2} \times (3)= (4)

Verify this conclusion against a fourth species involved- \text{N}_2 \; (g) for instance. Nitrogen isn't present in the net equation. The sum of its coefficient shall, therefore, be zero.

a + b = -1/2 + 1/2 = 0

Apply the Hess's Law based on the coefficients to find the enthalpy change of the last equation.

\Delta H _{(4)} = -\frac{1}{2} \; \Delta H _{(1)} + \frac{1}{2} \; \Delta H _{(2)} - \frac{1}{2} \; \Delta H _{(3)}\\\phantom{\Delta H _{(4)}} = -\frac{1}{2} \times (-628.9)+ \frac{1}{2} \times (-92.2) - \frac{1}{2} \times (184.7) \\\phantom{\Delta H _{(4)}} = 176.0 \; \text{kJ} \cdot \text{mol}^{-1}

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3 years ago
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Marrrta [24]

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Or can be expressed by converting kg to g.

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makvit [3.9K]

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Physical therapists commonly work in private workplaces and centers, clinics, patients' homes, and nursing homes. They invest quite a bit of their energy in their feet, effectively working with patients.

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