Answer:
Total mass of original sample = 0.38 g
Explanation:
Percentage of chloride = 74.5%
Mass of AgCl precipitate = 1.115 g
Mass of original sample = ?
Solution:
Mass of chloride:
1.115 g/ 143.3 g/mol × 35.5 g/mol
0.0078 g × 35.5 = 0.28 g
Total mass of compound:
Total mass = mass of chloride / %of Cl × 100%
Total mass = 0.28 g/ 74.5% × 100%
Total mass = 0.0038 g× 100
Total mass = 0.38 g
This problem is incomplete. Luckily, I found a similar problem from another website shown in the attached picture. The data given can be made to use through the Clausius-Clapeyron equation:
ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)
where
P₁ = 14 Torr * 101325 Pa/760 torr = 1866.51 Pa
T₁ = 345 K
P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K
ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
<em>ΔHvap = 48769.82 Pa/mol or 48.77 kPa/mol</em>
Answer:
See explanation
Explanation:
According to the law of conservation of mass; the total mass of reactants on the left hand side of the reaction equation is equal to the total mass of products on the right hand side of the reaction equation.
Hence, the total mass of each atom on either side of the reaction equation should be exactly the same.
Since there are two atoms of oxygen on the reactants side, the total mass of oxygen = 16 amu * 2 = 32 amu
Since there are two oxygen atoms on the products side, total mass of oxygen = 16 amu * 2 = 32 amu
Answer:
ratio=0.996m/s
Explanation:
; M= molecular weight of compound;
RMS speed is inversly proportional to the molecular weight hence compound having less molecular weight will have more rms speed value.


;
;
=ratio of
=
;


Molar mass NaCl = 58.44 g/mol
number of moles:
mass NaCl / molar mass
145 / 58.44 => 2.4811 moles of NaCl
Volume = 3.45 L
Therefore :
M = moles / volume in liters:
M = 2.4811 / 3.45
M = 0.719 mol/L⁻¹
hope this helps!