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2 years ago

Balance the equation C7H6O2 + O2 —> CO2 + H2O

Chemistry

1 answer:

ser-zykov [4K]2 years ago

4 0

2 C₇H₆O₂ + 15 O₂ → 14 CO₂ + 6 H₂O

Explanation:

C₇H₆O₂ + O₂ → CO₂ + H₂O

First we have to balance the O- atoms, we have to put 6 in front of water so there are 12 H atoms on RHS, to balance it we need to put 2 in front of C₇H₆O₂, and so we have 14 C - atoms on LHS, 28 + 6 = 34 O - atoms on RHS, so we have to put 15 in front of Oxygen in LHS, so that each and every atom in the equation gets balanced now. The balanced equation is, 2 C₇H₆O₂ + 15 O₂ → 14 CO₂ + 6 H₂O

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Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is init

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Answer:

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Explanation:

Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.

Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T}=k$

Where P = pressure, T = temperature, K = Constant

You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:

$\frac{P1}{T1}=\frac{P2}{T2}$

In this case:

P1= 1.50 atm
T1= 22 °C= 295 °K (being 0°C= 273 °K)
P2= ?
T2= 11 °C= 284 K

Replacing:

$\frac{1.5 atm}{295 K}=\frac{P2}{284 K}$

Solving:

$P2= 284 K*\frac{1.5 atm}{295 K}$

P2=1.44 atm

The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)

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