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Nezavi [6.7K]
2 years ago
7

Under what conditions of temperature and pressure is a gas least soluble in water

Chemistry
1 answer:
snow_tiger [21]2 years ago
7 0

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.

To know more about solubility of gas, visit the below link:

brainly.com/question/14747303

#SPJ4

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Which of these is caused by a chemical change?
IRISSAK [1]
I think that is A is the answer.
8 0
3 years ago
Americium-241 is used in smoke detectors. it has a first order rate constant for radioactive decay of k=1.6×10−3yr−1. by contras
Helga [31]
This question is missing the part that actually asks the question. The questions that are asked are as follows:

(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.

(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.

We can use the equation for a first order rate law to find the amount of material remaining after 4 days:

[A] = [A]₀e^(-kt)

[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.

(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.

4 days x 1 year/365 days = 0.0110

A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg

The decay of americium is so slow that no noticeable change occurs over 4 days.

(b) We can simply plug in the information of iodine-125 and solve for A:

A = (1.00)e^(-0.011 x 4)
A = 0.957 mg

Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.
7 0
4 years ago
Assuming 80 mg of vitamin c present 29.57 ml if oj how many milliliters of 0
BigorU [14]
I think you add 29.57 + 80 and the answer would be 30.37
5 0
3 years ago
How do the properties of compounds often compare with the properties of the elements that make them
const2013 [10]
The properties of a compound are nothing like the properties of th elements that combine to make it.


sodium chloride,NaCl, table salt, is a white cubic crystal, stable ionic compound, soluble in water
sodium is a soft metal, very reactive with water and air
chlorine is a yellow green gas that is poisonous and very reactive.

carbon dioxide, CO2, is a clear, colorless gas that puts out fires, plants use but we breath out.
carbon is a black solid, or a diamond
oxygen is a clear colorless gas that is needed for burning and breathing.
8 0
3 years ago
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
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