Answer:
a) 8×10²³ molecules of N2
Explanation:
Let us consider the same temperature and pressure of both sample are standard temperature and pressure.
Standard temperature = 273.15 K
Standard pressure = 1 atm
Number of moles of CO₂ = 0.107 mol
Volume of CO₂ = ?
Solution:
PV = nRT
R = general gas constant = 0.0821 atm.L /mol.K
1 atm × V = 0.107 mol × 0.0821 atm.L /mol.K × 273.15 K
1 atm × V = 2.4 atm.L
V = 2.4 atm.L/1 atm
V = 2.4 L
Volume of N₂:
Number of moles of N₂:
1 mole = 6.022 ×10²³ molecules
8×10²³ molecules × 1 mol / 6.022 ×10²³ molecules
1.33 mol
PV = nRT
R = general gas constant = 0.0821 atm.L /mol.K
1 atm × V = 1.33 mol × 0.0821 atm.L /mol.K × 273.15 K
1 atm × V = 29.83 atm.L
V = 29.83 atm.L/1 atm
V = 29.83 L
According to its formula FeSO4.7H2O
we can get the percent % by the mass of H2O from this formula
%mass of H2O = (mass of water H2O/ mass of the hydrate)x100
when the mass of water = molar mass x 7 = 18 x 7 = 126
and the mass of hydrate (feSO4) = molar mass = 278
So by substitution:
%mass of H2O = (126/278) x 100 = 45%
Answer:
Explanation:
You did not provide the reaction. However, you should know that only change in temperature affects the value of an equilibrium constant (Keq) by the equation, K=Ae>-RT
Answer:
B. It contains 15.0 grams of HNO₃ per 100 grams of solution.
Explanation:
- The percent of a solute in a solution is the mass of the solute (g) per 100 g of solution.
<em>So, an aqueous solution that is 15.0 percent HNO₃ by mass means that It contains 15.0 grams of HNO₃ per 100 grams of solution.</em>
Three of the intermolecular forces of attraction are roughly equal, the substances<span> will be </span>soluble in each other, <span>This means that </span>ionic<span> or </span>polar<span> solutes </span>dissolve<span> in </span><span>polar solvents</span>
