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.
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Answer:
The resulting solution contains approximately 666 g of water.
Explanation:
In the initial solution we have:
1g salt : 8g sugar : 200g water
This means that the ratios are:
In the final solution we have:
5g salt: xg sugar: yg water
The new ratios are:
Now we can calculate the amount of sugar in the final solution:
Finally, we calculate the amount of water:
Two or more compounds that have same molecular formula but differ in the arrangement of atoms in molecule and thus posses different properties are known as isomers.
The molecular formula of pentane is substituting one hydrogen from pentane with bromine results in the formation of monobromo derivatives of pentane having molecular formula,
.
The structure of monobromo derivatives of pentane that is 1-bromopentane, 2-bromopentane, and 3-bromopentane and having molecular formula, is shown in the image.
The other two arrangements of monobromo derivatives of pentane that is 2-bromo-2-methylbutane and 2-bromo-3-methylbutane is shown in the image.
There are different structures of monobromo derivatives of pentane having molecular formula, which contain a 4-carbon chain are 1- bromo-2-methylbutane and 1-bromo-3-methylbutane shown in the image.
