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Kobotan [32]
3 years ago
14

he solubility of oxygen gas in water at 25 °C and 1.0 atm pressure of oxygen is 0.041 g/L. The solubility of oxygen in water at

2.5 atm and 25 °C is ________ g/L.
Chemistry
1 answer:
Tema [17]3 years ago
8 0

Answer:

0.1025 g/L

Explanation:

From the question above,

The solubility of oxygen gas in water at 25°c and 1.0 atm pressure is 0.041g/L

Henry's law states at a constant temperature the solubility of a gas is proportional to the pressure

Since the temperatures are both similar (25°c) then, the solubility of oxygen in water at 2.5 atm can be calculated as follows

= 0.041/1.0 × 2.5

= 0.041×2.5

= 0.1025 g/L

Hence the solubility of oxygen in water at 2.5 atm and 25°c is 0.1025 g/L

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3 years ago
How many moles are in 35.6 g of H2O
klasskru [66]

Answer:

Given

mass of H2O (m) =35.6g

molarmass (mr) = H2O ), 1x2+16=18g/mol

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3 0
3 years ago
An experiment shows that a 250 ?ml gas sample has a mass of 0.430 g at a pressure of 736 mmhg and a temperature of 28 ?c.
dmitriy555 [2]
What we're looking for here is the gas sample's molar mass given its mass, pressure, volume, and temperature. Recalling the gas law, we have

PV = nRT or
n = \frac{PV}{RT}

where R is <span>0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.

Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R. 
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Thus, we have

P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L

Now, applying these converted values into the gas law, we have

n = \frac{(0.968 atm)(0.250 L)}{(0.08206 L.atm/mol.K)(301.15 K)}
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Given that the mass of the sample is 0.430 g, we have

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Name the mineral group for each of the following minerals: Group of answer choices Kyanite (Al2SiO5) Ilmenite (FeTiO3) Rhodochro
Ainat [17]

Answer:

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Ilmenite (FeTiO3) - Oxides

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Explanation:

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