The answer is C. It will depend on the speed.
Answer: 98.36g/mol
Explanation:Please see attachment for explanation
Data:
<span>Solute: 28.5 g of glycerin (C3H8O3)
Solvent: 135 g of water at 343 k.
Vapor pressure of water at 343 k: 233.7 torr.
Quesiton: Vapor pressure of water
Solution:
Raoult's Law: </span><span><span>The vapour
pressure of a solution of a non-volatile solute is equal to the vapour
pressure of the pure solvent at that temperature multiplied by its mole
fraction.
Formula: p = Xsolvent * P pure solvent
X solvent = moles solvent / moles of solution
molar mass of H2O = 2*1.0g/mol + 16.0 g/mol = 18.0 g/mol
moles of solvent = 135 g of water / 18.0 g/mol = 7.50 mol
molar mass of C3H8O3 = 3*12.0 g/mol + 8*1 g/mol + 3*16g/mol = 92 g/mol
moles of solute = 28.5 g / 92.0 g/mol = 0.310 mol
moles of solution = moles of solute + moles of solvent = 7.50mol + 0.310mol = 7.810 mol
Xsolvent = 7.50mol / 7.81mol = 0.960
p = 233.7 torr * 0.960 = 224.4 torr
Answer: 224.4 torr
</span> </span>
Answer:
0.676 grams of manganese (IV) oxide should be added.
Explanation:
Moles of chlorine gas = n
Volume of the chlorine gas = V = 205 mL = 0.205 L
Pressure of the chlorine gas = 705 Torr = 
1 atm = 760 Torr
Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K
( ideal gs equation)
According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :
of manganese (IV) oxide
Mass of 0.00777 moles of manganese (IV) oxide:
0.00777 mol × 87 g/mol = 0.676 g
0.676 grams of manganese (IV) oxide should be added.
