The molarity of the diluted solution is 0.32 M
Considering the question given above, the following data were obtained:
Volume of stock solution (V₁) = 500 mL
Molarity of stock solution (M₁) = 2.1 M
Volume of diluted solution (V₂) = 3.25 L = 3.25 × 1000 = 3250 mL
<h3>Molarity of diluted solution (M₂) =....? </h3>
The molarity of the diluted solution can be obtained as follow:
<h3>M₁V₁ = M₂V₂</h3>
2.1 × 500 = M₂ × 3250
1050 = M₂ × 3250
<h3>Divide both side by 3250</h3><h3 />
M₂ = 1050 / 3250
<h3>M₂ = 0.32 M</h3>
Therefore, the molarity of the diluted solution is 0.32 M
Learn more: brainly.com/question/22325751
The keg for the reaction
2 SO2(g) + O2(g) → 2 SO3(g) is
Keg = [SO3]^2/ {(SO2)^2 ( O2)}
Keg (equilibrium constant) is the ratio of of equilibrium concentration of the product raised to the power of their stoichiometric coefficient to the equilibrium concentration of the reactant raised to the power of their stoichiometric coefficient.
<span>A) elements and compounds.
Compounds are two combined elements.</span>
Answer:
1 mol
Explanation:
Using the general gas law equation as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the provided information in the question;
V = 22.4L
T = 273K
P = 1 atm
R = 0.0821 Latm/molK
n = ?
Using PV = nRT
n = PV/RT
n = (1 × 22.4) ÷ (0.0821 × 273)
n = 22.4 ÷ 22.4
n = 1mol
