Answer : The normal boiling point of ethanol will be, 
or 
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of ethanol at 
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the normal boiling point of ethanol will be, or
Answer:
1.414 Moles
Solution:
Data Given:
Mass of MgS₂O₃ = 193 g
M.Mass of MgS₂O₃ = 136.43 g.mol⁻¹
Moles = ?
Formula Used:
Moles = Mass ÷ M.Mass
Putting values,
Moles = 193 g ÷ 136.43 g.mol⁻¹
Moles = 1.414 mol
To take the percent by mass of this element, we use the
formula:
% mass = (mass of element / mass of ore) * 100%
% mass = (47.5 g / (660 kg * 1000 g / kg)) * 100*
<span>% mass = 7.20 x 10^-3 %</span>
1.38 moles of oxygen
Explanation:
Thermal decomposition of Lead (II) nitrate is shown by the balanced equation below;
2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
The mole ration of Lead (II) nitrate to oxygen is 2: 1
Therefore 2.76 moles of Lead (II) nitrate will lead to production of? moles of oxygen;
2: 1
2.76: x
Cross-multiply;
2x = 2.76 * 1
x = 2.76 / 2
x = 1.38
Answer:
Hi! I believe this is your answer:
52 kilograms
Hope this helps, sorry if it's wrong!
Explanation:
