Answer is: 0,0030 mol of carbon dioxide.
Carbon dioxide solubility in water at 20°C and 1 atm is: 3,8·10⁻² mol/L.
Carbon dioxide solubility in water at 25°C and 1 atm is: 3,5·10⁻² mol/L.
Difference is: 0,038 mol/L - 0,035 mol/L = 0,003 mol/L.
V(carbon dioxide) = 1 L.
n(carbon dioxide) = V · c.
n(carbon dioxide) = 1 L · 0,003 mo/L.
n(carbon dioxide) = 0,0030 mol.
1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution
=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution
2) Molarity = M = number of moles of solute / liters of solution
=> calculate the number of moles of 8.9 grams of NaCl
3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol
4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol
5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M
Answer: 0.152 M
the balanced chemical equation for the decomposition of H₂O₂ is as follows
2H₂O₂ ---> 2H₂O + O₂
stoichiometry of H₂O₂ to O₂ is 2:1
the number of moles of H₂O₂ decomposed is - 0.250 L x 3.00 mol/L = 0.75 mol
according to stoichiometry the number of O₂ moles is half the number of H₂O₂ moles decomposed
number of moles of O₂ - 0.75 mol / 2 = 0.375 mol
apply the ideal gas law equation to find the volume
PV = nRT
where P - standard pressure - 10⁵ Pa
V - volume
n - number of moles 0.375 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - standard temperature - 273 K
substituting the values in the equation
10⁵ Pa x V = 0.375 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 8.5 L
volume of O₂ gas is 8.5 L
Answer:
0.019 moles of M2CO3
Explanation:
M2CO3(aq) + BaCl2 (aq) --> 2MCl (aq) + BaCO3(s)
From the equation above;
1 mol of M2CO3 reacts to produce 1 mol of BaCO3
Mass of BaCO3 formed = 3.7g
Molar mass of BaCO3 = 197.34g/mol
Number of moles = Mass / Molar mass = 3.7 / 197.34 = 0.0187 ≈ 0.019mol
Since 1 mol of M2CO3 reacts with 1 mol of BaCO3,
1 = 1
x = 0.019
x = 0.019 moles of M2CO3
<span>The molar mass of the compound is 122 g. </span>
