The conditions of temperature and pressure in which a gas least soluble in water is low pressure and high temperature.
<h3>What is Henry Law?</h3>
The amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid, according to Henry's law.
From this law it is clear that:
- As the pressure of the gas increases solubility of the gas on the liquid also increases.
But if the temperature of the liquid decreases then the solubility of the gas also increases.
Hence at low pressure and high temperature, gas is least soluble.
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Remember that any intersection of lines is a C, and that the number of hydrogens attached are the necessary to complet the 4 bonds.
1) CH3 - CH (OH) - CH (CH3) -CH3
2) CH3 - O - CH(CH3)-CH2 - CH3
I have used the parenthesis to indicate that the radical inside is in other branch, bonded by a single line -
Answer:
Basically, all phosphates except Sodium phosphates, Potassium phosphates and Ammonium phosphates are insoluble in water. That, of course, includes Magnesium phosphate.
Explanation:
Hope this helped!
The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane(
) = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane(
) = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
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Answer:
The answer is
<h2>0.052 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula

From the question
mass = 2.5 g
volume = 48 cm³
The density is

We have the final answer as
<h3>0.052 g/cm³</h3>
Hope this helps you