Answer:
K remains the same;
Q < K;
The reaction must run in the forward direction to reestablish the equilibrium;
The concentration of
will decrease.
Explanation:
In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.
Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.
The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.
As a result, since
is also a reactant, its concentration will decrease.
Answer:
Coefficient
Explanation:
Chemical equation:
aA + bB → dD
This equation shows that reaction is balanced by coefficient q,b and d.
Reactant coefficients Product Coefficient
A a D d
B b
Every balanced equation follow the law of conservation of mass.
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
Explanation:
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
For example:
In given photosynthesis reaction:
6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂
there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass. The numbers 6 with CO₂, 6 with H₂O and 6 with O₂ are coefficient which balance the equation.
Answer:
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Explanation:
Answer:
A. 30cm³
Explanation:
Based on the chemical reaction:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
<em>1 mol of calcium carbonate reacts with 2 moles of HCl to produce 1 mol of CO₂</em>
<em />
To solve this question we must convert the mass of each reactant to moles. With the moles we can find limiting reactant and the moles of CO₂ produced. Using PV = nRT we can find the volume of the gas:
<em>Moles CaCO₃ -Molar mass: 100.09g/mol-</em>
1.00g * (1mol / 100.09g) = 9.991x10⁻³ moles
<em>Moles HCl:</em>
50cm³ = 0.0500dm³ * (0.05 mol / dm³) = 2.5x10⁻³ moles
For a complete reaction of 2.5x10⁻³ moles HCl there are necessaries:
2.5x10⁻³ moles HCl * (1mol CaCO₃ / 2mol HCl) = 1.25x10⁻³ moles CaCO₃. As there are 9.991x10⁻³ moles, HCl is limiting reactant.
The moles produced of CO₂ are:
2.5x10⁻³ moles HCl * (1mol CO₂ / 2mol HCl) = 1.25x10⁻³ moles CO₂
Using PV = nRT
<em>Where P is pressure = 1atm assuming STP</em>
<em>V volume in L</em>
<em>n moles = 1.25x10⁻³ moles CO₂</em>
<em>R gas constant = 0.082atmL/molK</em>
<em>T = 273.15K at STP</em>
<em />
V = nRT / P
1.25x10⁻³ moles * 0.082atmL/molK*273.15K / 1atm = V
0.028L = V
28cm³ = V
As 28cm³ ≈ 30cm³
Right option is:
<h3>A. 30cm³</h3>