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3 years ago

A mixture of helium, nitrogen, and oxygen has a total pressure of 752 mm Hg. The

Chemistry

1 answer:

NISA [10]3 years ago

5 0

Answer: The partial pressure of oxygen in the mixture is 321 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_{He}+p_{N_2}+p_{O_2}$

Given : $p_{total}$ = total pressure of gases = 752 mm Hg

$p_{He}$ = partial pressure of Helium = 234 mm Hg

$p_{N_2}$ = partial pressure of nitrogen = 197 mm Hg

$p_{O_2}$ = partial pressure of oxygen = ?

Putting in the values we get:

$752mmHg=234mmHg+197mmHg+p_{O_2}$

$p_{O_2}=321mmHg$

The partial pressure of oxygen in the mixture is 321 mm Hg

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Identify the correct coefificients to balance it -C3H8+O2 to -CO2 +-H2O

Iteru [2.4K]

Answer:

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Explanation:

${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Among the four species in this reaction, $\rm C_3H_8$ is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is $1$ .

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Number of atoms on the left-hand side of the reaction:

$\rm C$ : $1 \times 3 = 3$ .
$\rm H$ : $1 \times 8 = 8$ .
$\rm O$ : not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

$3 / 1 = 3$ $\rm CO_2$ molecules, and
$8 / 2 = 4$ $\rm H_2O$ molecules.

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Number of atoms on the right-hand side of the reaction:

$\rm C$ : $3 \times 1 = 3$ .
$\rm H$ : $4 \times 2 = 8$ .
$\rm O$ : $3 \times 2 + 4 \times 1 = 10$ .

The number of $\rm O$ atoms on the left-hand side should match those on the right-hand side. In this reaction, $\rm O_2$ is the only reactant with $\rm O\!$ atoms. These $10$ $\rm \! O$ atoms would correspond to: