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natulia [17]
3 years ago
13

Consider the chemical equation for the combustion of sugar. _C6H12O6(s) + _O2(g) Right arrow. _CO2(g) + _H2O(l) Which sequence o

f coefficients should be placed in the blanks to balance this equation?
Chemistry
2 answers:
Alisiya [41]3 years ago
8 0

Answer:

1666

Explanation:

ioda3 years ago
6 0

Answer:

1C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)

Explanation:

In general equation combustions where all carbon produce CO₂ and all hydrogen H₂O we must begin the balance with these two species:

_C₆H₁₂O₆(s) + _O₂(g) → _CO₂(g) + _H₂O(l)

If sugar is 1, CO₂ and H₂O must be:

1C₆H₁₂O₆(s) + _O₂(g) → 6CO₂(g) + 6H₂O(l)

Thus, C and H are balanced. In the right side we have 12 oxygen from CO₂ + 6 oxygen from H₂O = 18 oxygen

In the left there are 6, thus, O₂ must provide 12 oxygens, thus:

<h3>1C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)</h3>

And this is the balanced reaction

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A sample of methane collected when the temp was 30 C mmHg measures 398 mL. What would be the volume of the sample at -5 C and 61
Levart [38]
<h2>Question </h2>

A sample of methane collected when the temp was 30 C and 760mmHg measures 398 mL. What would be the volume of the sample at -5 C and 616 mmHg pressure

<h2>Answer:</h2>

434.32mL

<h2>Explanation:</h2>

Using the combined gas law:

\frac{PV}{T} = k

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T = Temperature

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\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = k           ---------------------(i)

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V₁ and V₂ are the initial and final volumes of the given gas

T₁ and T₂ are the initial and final temperatures of the gas.

<em>From the question:</em>

the gas is methane

P₁  = 760mmHg

P₂ = 616mmHg

V₁ = 398mL

V₂ = ?

T₁ = 30°C = (30 +273)K = 303K

T₂ = -5°C = (-5 +273)K = 268K

Substitute these values into equation (i) as follows;

\frac{760*398}{303} = \frac{616*V_2}{268}

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V₂ = \frac{760*398*268}{616*303}

V₂ = 434.32mL

Therefore, the volume of the sample at -5C and 616mmHg pressure is 434.32mL

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