Suppose 110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen fluoride dissolves in water to form 150.0 mL of an aqueous solution. 0.032 M is the concentration of the resulting hydrofluoric acid.
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now write the balanced chemical equation
H₂ + F₂ → 2HF
<h3>What is Ideal Gas ?</h3>
An ideal gas is a gas that obey gas laws at all temperature and pressure conditions. It have velocity and mass but do not have volume. Ideal gas is also called perfect gas. Ideal gas is a hypothetical gas.
It is expressed as:
PV = nRT
where,
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant
T = temperature
Here,
P = 1 atm [At STP]
V = 110 ml = 0.11 L
T = 273 K [At STP]
R = 0.0821 [Ideal gas constant]
Now put the values in above expression
PV = nRT
1 atm × 0.11 L = n × 0.0821 L.atm/ K. mol × 273 K

n = 0.0049 mol
<h3>How to find the concentration of resulting solution ? </h3>
To calculate the concentration of resulting solution use the expression

= 0.032 M
Thus from the above conclusion we can say that Suppose 110.0 mL of hydrogen gas at STP combines with a stoichiometric amount of fluorine gas and the resulting hydrogen fluoride dissolves in water to form 150.0 mL of an aqueous solution. 0.032 M is the concentration of the resulting hydrofluoric acid.
Learn more about the Ideal Gas here: brainly.com/question/25290815
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Answer:
Air is a mixture. Its constituents can be separated. For example: oxygen, nitrogen etc.
Hydrochloric acid is a mixture, being an acidic liquid.
The answer is A. Solids only
Answer:
5 L
Explanation:
We'll begin by calculating the molarity of the CaCl₂ solution. This can be obtained as follow:
Mole of CaCl₂ = 0.5 mole
Volume = 2 L
Molarity =?
Molarity = mole /Volume
Molarity = 0.5 / 2
Molarity = 0.25 M
Finally, we shall determine the volume of the diluted solution. This can be obtained as follow:
Molarity of stock solution (M₁) = 0.25 M
Volume of stock solution (V₁) = 2 L
Molarity of diluted solution (M₂) = 0.1 M
Volume of diluted solution (V₂) =?
M₁V₁ = M₂V₂
0.25 × 2 = 0.1 × V₂
0.5 = 0.1 × V₂
Divide both side by 0.1
V₂ = 0.5 / 0.1
V₂ = 5 L
Thus the volume of the diluted solution is 5 L