C = 0.11 mol
V = 5.65 L
n = ???
n = C*V
n = 0.11 * 5.65
n = 0.622 mols
1 mol of CaCl2 = 40 + 2*35.5 = 111 grams
0.622 mol = x
x = 111 * 0.622
x = 69.0 grams CaCl2
Answer is: 9,7 L is needed to store helium gas.
n(He) = 0,80 mol.
p(He) = 204,6 kPa.
T = 300 K.
R = 8,314 J/K·mol; universal gas constant.
Use ideal law eqaution: p·V = n·R·T.
V = n·R·T / p.
V(He) = 0,80 mol · 8,314 J/K·mol · 300 K ÷ 204,6 kPa.
V(He) = 9,75 L.
Answer:
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Explanation:
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Answer:
The correct answer is 0.6 mL.
Explanation:
We use the mathematical expression:
Ci x Vi = Cf x Vf
Where Ci is the initial concentration (5 M); Cf and Vf refers to final concentration (0.002 M) and final volume (1500 mL). With the given data, we calculate the initial volume (Vi):
Vi = (Cf x Vf)/Ci = (0.002 M x 1500 mL)/(5 M) = 0.6 mL
Therefore, we need 0.6 mL of 5 M NaCl to prepare the solution with the requested dilution.
