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2 years ago

The mass of a quantity of Cl gas (70,906 g/mol) that occupies 50.0L at 27.0°C and 721 mm Hg is:

Chemistry

1 answer:

inessss [21]2 years ago

4 0

We need no of moles

Apply ideal gas equation

PV=nRT
n=PV/RT
n=721(50)/8.314(300)
n=14.4mol

mass

70.906(14.4)
1021g

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3 years ago

Calculate the morality of a 35.4% (by mass) aqueous solution of phosphoric acid (H3PO4). Molar mass of H3PO4 is 98.00g/mol

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2 years ago

Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

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2 years ago

For the hypothetical reaction: A + B ---> 2C, write an expression that relates the disappearance of A and B to the appearance

forsale [732]

Answer:

Rate expression has been given below

Explanation:

According to the given equation, 1 molecule of A reacts with 1 molecule of B and produces 2 molecules of B at a time.

So, rate of disappearance of both A and B are one half of rate of appearance of B

Hence rate expression can be represented as:

$Rate=\frac{-\Delta [A]}{\Delta t}=\frac{-\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}$

where $\frac{-\Delta [A]}{\Delta t}$ is rate of disappearance of A, $\frac{-\Delta [B]}{\Delta t}$ is rate of disappearance of B and $\frac{\Delta [C]}{\Delta t}$ rate of appearance of C