Answer:
The concentration of the CaBr2 solution is 96 µmol/L
Explanation:
<u>Step 1:</u> Data given
Moles of Calciumbromide (CaBr2) = 4.81 µmol
Volume of the flask = 50.0 mL = 0.05 L
<u>Step 2:</u> Calculate the concentration of Calciumbromide
Concentration CaBr2 = moles CaBr2 / volume
Concentration CaBr2 = 4.81 µmol / 0.05 L
Concentration CaBr2 = 96.2 µmol /L = 96.2 µM
The concentration of the CaBr2 solution is 96 µmol/L
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Answer:
The answer to your question is 8.74 g of He
Explanation:
Data
V = 2.4 x 10² L
P = 99 kPa
T = 0°C
mass = ?
Process
1.- Convert kPa to atm
P = 99 kPa = 99000 Pa
1 atm --------------- 101325 Pa
x --------------- 99000 Pa
x = (99000 x 1) / 101325
x = 0.977 atm
2.- Convert temperature to °K
°K = 273 + 0
°K = 273
3.- Substitution
PV = nRT
- Solve for n
n = PV / RT
n = (0.977)(2.4 x 10²) / (0.082)(273)
n = 24.48 / 22.386
n = 1.093 moles
4.- Calculate the grams of He
8 g -------------------- 1 mol
x -------------------- 1.093 moles
x = (1.093 x 8) / 1
x = 8.74 g