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lilavasa [31]
3 years ago
11

) determine the henry's law constant for ammonia in water at 25°c if an ammonia pressure of 0.022 atm produces a solution with a

concentration of 1.3 m.
a.59 m/atm

b.0.029 m/atm

c.0.038 m/atm

d.35 m/atm

e.0.017 m/atm
Chemistry
1 answer:
Nataly [62]3 years ago
3 0

Answer:

a. 59 m/atm

Explanation:

  • To solve this problem, we must mention Henry's law.
  • <em>Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.</em>
  • It can be expressed as: C = KP,

C is the concentration of the solution (C = 1.3 M).

P is the partial pressure of the gas above the solution (P = 0.022 atm).

K is the Henry's law constant (K = ??? M/atm),

∵ C = KP.

∴ K = C/P = (1.3 M)/(0.022 atm) = 59.0 M/atm.

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Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76
aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

= 0.0144 moles

The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

moles = 0.461 g /16 g/mol

= 0.0288 moles

The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

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3 years ago
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