Answer:
442.3 mL
Explanation:
Remember that Molarity is a measure of concentration in Chemistry and it's defined as the number of moles of the substance divided by liters of the solution:

Then, you can express 11.27 g of AgNO3 as moles of AgNO3 using the molar mass of the compound:

Then you can solve for the volume of the solution:

Hope it helps!
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M
Answer:
81.5 L
Explanation:
We can use the combined gas law equation that gives the relationship among pressure, temperature and volume of gases for a fixed amount of gas.
P1V1 / T1 = P2V2 / T2
where P1 - pressure, V1 - volume and T1 - temperature at the first instance
P2 - pressure, V2 - volume and T2 - temperature at the second instance
substituting the values in the equation
1240 Torr x 47.2 L / 298 K = 730 Torr x V2 / 303 K
V2 = 81.5 L
the new volume the gas would occupy when the conditions have changed is 81.5 L
According to the illustration, the vanadium (V) oxide would be a catalyst.
<h3>What are catalysts?</h3>
Catalysts are substances that are utilized in reactions that are not themselves consumed in reactions but only speed up the rate of the reactions.
Catalysts speed up the rate of reactions by lowering the activation energy of the reactants.
Sulfur dioxide reacts with oxygen to produce sulfur trioxide. The vanadium (v) oxide is not consumed in the reaction. Thus it only serves as a catalyst.
More on catalysts can be found here: brainly.com/question/12260131
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Molality is one way of expressing concentration for solutions. It has units of moles of solute per kg of solvent. From the given values, we easily calculate for the moles of solute by multiplying the mass of solvent to the molality. We do as follows:
moles solute = 0.3 (10) = 3 mol solute