Answer:
Mass = 121 g
Explanation:
Given data:
Mass in gram of CO₂ = ?
Volume = 61.8 L
Pressure = standard = 1 atm
Temperature = 273.15 K
Solution:
Formula:
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
1 atm × 61.8 L = n ×0.0821 atm.L/ mol.K × 273.15 k
61.8 L.atm = 22.42 atm.L/ mol × n
n = 61.8 L.atm /22.42 atm.L/ mol
n = 2.76 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 2.76 mol × 44 g/mol
Mass = 121 g
For the answer to the question above, well presumably because the exact concentration of the composition KMnO4 solution doesn't matter. <span>If the concentration of the KMnO4 solution is important (usually in titrations etc.) then it is not allowed to use a wet bottle. The water in the bottle will dilute the KMnO4 solution and change the concentration of the said compound.</span>
Calcium Floride (Caf2)
Hope this helped =D
a. 34 mL; b. 110 mL
a. A tablet containing 150 Mg(OH)₂
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
<em>Moles of Mg(OH)₂</em> = 150 mg Mg(OH)₂ × [1 mmol Mg(OH)₂/58.32 mg Mg(OH)₂
= 2.572 mmol Mg(OH)₂
<em>Moles of HCl</em> = 2.572 mmol Mg(OH)₂ × [2 mmol HCl/1 mmol Mg(OH)₂]
= 5.144 mmol HCl
Volume of HCl = 5.144 mmol HCl × (1 mmol HCl/0.15 mmol HCl) = 34 mL HCl
b. A tablet containing 850 mg CaCO₃
CaCO₃ + 2HCl ⟶ CaCl₂ + CO₂ + H₂O
<em>Moles of CaCO₃</em> = 850 mg CaCO₃ × [1 mmol CaCO₃/100.09 mg CaCO₃
= 8.492 mmol CaCO₃
<em>Moles of HCl</em> = 8.492 mmol CaCO₃ × [2 mmol HCl/1 mmol CaCO₃]
= 16.98 mmol HCl
Volume of HCl = 16.98 mmol HCl × (1 mL HCl/0.15 mmol HCl) = 110 mL HCl